Class 12 Physics (Part 2) Chapter 16 Refraction of Light at Spherical Surfaces : Lenses

Key Concept: Refraction at Concave Spherical Surface, Small-Angle Approximation

d) $\beta = \gamma - \alpha$
[Solution Description]
From the geometry of the concave surface and the small-angle approximation:
$\alpha = \tan \alpha = \frac{h}{u}$
$\beta = \tan \beta = \frac{h}{v}$
$\gamma = \tan \gamma = \frac{h}{R}$
The correct relation derived for small angles is:
$n_1 (\gamma - \alpha) = n_2 (\gamma - \beta)$
However, the question asks specifically about the geometric relations.

Key Concept: Refraction at a Spherical Surface, Combined Concepts

b) $119$ cm, real
[Solution Description]
First, apply refraction at the first convex surface ($R_1 = +20$ cm, $n_1 = 1$, $n_2 = 1.5$, $u = -60$ cm):
$\frac{1.5}{v_1} - \frac{1}{-60} = \frac{1.5 - 1}{20}$
$\frac{1.5}{v_1} + 0.0167 = 0.025$
$\frac{1.5}{v_1} = 0.0083 \implies v_1 = 180$ cm$
The image from the first surface becomes the object for the second surface. Distance between surfaces is negligible, so $u_2 = -180$cm (virtual object for the second surface). For the second convex surface ($R_2 = -30$cm,$n_1 = 1.5$,$n_2 = 1$): \\ $ \\ \frac{1}{v_2} - \frac{1.5}{-180} = \frac{1 - 1.5}{-30}$
$\frac{1}{v_2} + 0.0083 = 0.0167$
$\frac{1}{v_2} = 0.0084 \implies v_2 \approx 119$ cm
$
The final image is real and located 119 cm to the right of the second surface.

Key Concept: Focal Length Calculation

c) 20 cm
[Solution Description]
Using the lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Given $u = -30$ cm (by sign convention) and $v = 60$ cm, substitute these values:
$\frac{1}{f} = \frac{1}{60} - \frac{1}{-30} = \frac{1}{60} + \frac{1}{30} = \frac{3}{60} = \frac{1}{20}$
Therefore, $f = 20$ cm.

Key Concept: Gaussian Form of Lens Equation

c) 13.33 cm
[Solution Description]
The Gaussian form of the lens equation is:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given $u = -40 \, \text{cm}$ (object distance is negative as per sign convention) and $f = 10 \, \text{cm}$, substitute into the equation:
$\frac{1}{v} - \frac{1}{-40} = \frac{1}{10}$
This simplifies to:
$\frac{1}{v} = \frac{1}{10} - \frac{1}{40}$
$\frac{1}{v} = \frac{4 - 1}{40} = \frac{3}{40}$
Therefore, the image distance $v = \frac{40}{3} \approx 13.33 \, \text{cm}$.

Key Concept: Concave Lenses and Focal Length

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a plano-concave lens has one flat surface and one concave surface, which is correct. It also claims the focal length of such a lens is always negative, which is true because plano-concave lenses are diverging lenses. The reason states that all diverging lenses have a negative focal length, which is also correct. Moreover, the reason correctly explains why the assertion is true—because plano-concave lenses are diverging lenses, their focal lengths must be negative. Therefore, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Chromatic Aberration

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Chromatic aberration occurs in lenses because the refractive index ($n$) of the lens material changes with the wavelength (\$ \lambda \$) of light, causing dispersion. This variation is described by Cauchy's formula: $n = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + \dots$ On the other hand, mirrors do not exhibit chromatic aberration because reflection is independent of the wavelength of light. Thus, both Assertion and Reason are true, and Reason correctly explains why chromatic aberration occurs in lenses but not in mirrors.

Key Concept: Linear magnification and nature of image

b) $-10$ cm
[Solution Description]
Given:
$u = -20$ cm (object distance, negative as per sign convention),
$m = +0.5$.
For a concave lens, the magnification formula is:
$m = \frac{v}{u}.$
Substituting the known values:
$0.5 = \frac{v}{-20}.$
Solving for $v$:
$v = 0.5 \times (-20) = -10 \text{ cm.}$
Negative $v$ implies the image is virtual and on the same side as the object.

Key Concept: Principal Axis

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that for a plano-convex lens, the principal axis passes through the optical centre and the centre of curvature of the curved surface. This is true because the principal axis is defined as the line joining the optical centre and the centres of curvature of both surfaces. For a plano-convex lens, the plane surface has an infinite radius of curvature, meaning its centre of curvature lies at infinity. Therefore, the principal axis only passes through the optical centre and the centre of curvature of the curved surface.
The reason states that in a plano-convex lens, one surface is plane, so its centre of curvature lies at infinity. This is also true because a plane surface can be considered as part of a sphere with an infinitely large radius, hence its centre of curvature is at infinity.
However, the reason does not explain why the principal axis passes through the optical centre and the centre of curvature of the curved surface; it merely states a property of the plane surface. Hence, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Image Formation by Thin Lenses, Ray Diagrams for Convex Lens, Lens Maker's Formula

d) Assertion is false, but Reason is true.
[Solution Description]
To verify the Assertion (A), first, use the Gaussian lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given: $f = 20\,\text{cm}$, $u = -30\,\text{cm}$ (negative as per sign convention). Substituting these values:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$
Simplify to solve for $v$:
$\frac{1}{v} + \frac{1}{30} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$
Thus, $v = 60\,\text{cm}$. Since $v > 0$, the image is real and situated beyond $2f$ (as $2f = 40\,\text{cm}$), which contradicts the Assertion that it lies between $f$ and $2f$.
The Reason (R) correctly states the Gaussian lens formula and that for $u > f$ in a convex lens, the image is real. However, the Assertion incorrectly describes the position of the image relative to $f$ and $2f$. Therefore, Assertion is false, but Reason is true.

Key Concept: Lens Combination, Power

b) +1.67 D
[Solution Description]
Step 1: Calculate power of convex lens ($P_1$). Since $P = \frac{1}{f}$ and focal length $f_1 = 20$ cm = 0.2 m, then $P_1 = \frac{1}{0.2} = +5$ D.
Step 2: Calculate power of concave lens ($P_2$). Focal length $f_2 = -30$ cm = -0.3 m (negative because it's concave), so $P_2 = \frac{1}{-0.3} = -3.\overline{3}$ D.
Step 3: Combined power $P = P_1 + P_2 = 5 + (-3.\overline{3}) = +1.\overline{6}$ D $\approx 1.67$ D.

Key Concept: Lens maker's formula, refractive index

d) Assertion is false, but Reason is true.
[Solution Description]
The lens maker's formula in a medium is given by:
$\frac{1}{f} = \left(\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Initially, the lens is in air ($n_{\text{medium}} = 1$), so:
$\frac{1}{20} = (1.5 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
When immersed in the liquid ($n_{\text{medium}} = 1.25$):
$\frac{1}{f'} = \left(\frac{1.5}{1.25} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.2 - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.4 \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Comparing with the initial equation, we see that $\frac{1}{f'} = 0.8 \times \frac{1}{20}$, so $f' = 25$ cm. Thus, the focal length increases, not decreases. Hence, the Assertion is false. However, the Reason correctly states the modified lens maker's formula when the lens is in a medium, but it does not support the Assertion. Therefore, the Reason is true.

Key Concept: Optical Centre, Image Formation

c) 60 cm
[Solution Description]
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ where $u = -20$ cm (real object) and $f = +15$ cm (convex lens). Solving:
$\frac{1}{v} = \frac{1}{15} + \frac{1}{-20} = \frac{4-3}{60} = \frac{1}{60}$. Thus $v = +60$ cm. The optical centre ratio doesn't affect image position for thin lenses.

Key Concept: Convex Lenses and Optical Centre

d) It goes straight without deviation or displacement.
[Solution Description]
The optical centre of a lens is the point where a ray passes undeviated and undisplaced. For a thin convex lens, any ray passing through the optical centre continues straight without any bending or shift.

Key Concept: Image Formation by Thin Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a convex lens, when the object is placed beyond the focal point, a real and inverted image is formed. When the object is placed between the focal point and the lens, a virtual and erect image is formed. Therefore, the Assertion is true as a convex lens can form both real and virtual images. The Reason is also true because the position of the object relative to the focal point determines whether the image is real or virtual. Moreover, the Reason correctly explains the Assertion, as the nature of the image directly depends on the object's position relative to the focal point.

Key Concept: Lens Maker's Formula - Double Convex Lens

b) 16.67 cm
[Solution Description]
For a double convex lens, the lens maker’s formula becomes:
$\frac{1}{f} = (n - 1) \left( \frac{2}{R} \right)$
Given: $n = 1.6$, $R = 20$ cm. Substituting these values:
$\frac{1}{f} = (1.6 - 1) \left( \frac{2}{20} \right) = 0.6 \times 0.1 = 0.06$
Taking reciprocal gives:
$f = \frac{1}{0.06} \approx 16.67 \text{ cm}$

Key Concept: Image Formation by Concave Lens

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion (A) states that for a concave lens, the image formed is always virtual and diminished, which is correct as per the syllabus. This is because a concave lens diverges incident rays, causing them to appear to come from a point on the same side as the object, resulting in a virtual, erect, and diminished image.
The Reason (R) explains that a concave lens diverges parallel rays so that they appear to originate from the second focus after refraction. This is also true as per the ray diagram description in the syllabus. However, while the Reason provides a property of the concave lens, it does not directly explain why the image is always virtual and diminished.
Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Refraction at Spherical Surfaces

c) -20 cm on the left side of the surface
[Solution Description]
Using the formula for refraction at a convex spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Here, $n_1 = 1$, $n_2 = 1.5$, $u = -10$ cm (object distance is negative as per sign convention), $R = +20$ cm.
Substituting values:
$\frac{1.5}{v} - \frac{1}{-10} = \frac{1.5 - 1}{20}$.
Simplifying:
$\frac{1.5}{v} + \frac{1}{10} = \frac{0.5}{20}$.
$\frac{1.5}{v} = \frac{1}{40} - \frac{1}{10} = -\frac{3}{40}$.
Solving for $v$: $v = \frac{-40 \times 1.5}{3} = -20$ cm.
The negative sign indicates the image is virtual and on the same side as the object.

Key Concept: Refraction at Spherical Surfaces, Lens Maker's Formula

b) 30 cm
[Solution Description]
Given $R_1 = 15$ cm, $R_2 = -30$ cm (since the second surface is concave to the incident light), $n_{\text{lens}} = 1.6$, and $n_{\text{medium}} = 1.2$. The relative refractive index is $n = \frac{n_{\text{lens}}}{n_{\text{medium}}} = \frac{1.6}{1.2} = \frac{4}{3}$. Using the lens maker's formula:
$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{4}{3} - 1 \right) \left( \frac{1}{15} - \frac{1}{-30} \right)$
Simplifying:
$\frac{1}{f} = \frac{1}{3} \left( \frac{1}{15} + \frac{1}{30} \right) = \frac{1}{3} \times \frac{3}{30} = \frac{1}{30}$
Therefore, $f = 30$ cm.

Key Concept: Refraction at Concave Surface

c) $-25.71$ cm
[Solution Description]
The refraction formula for a spherical surface is:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Here, $n_1 = 1$, $n_2 = 1.5$, $u = -30$ cm, and $R = -20$ cm. Substituting these values:
$\frac{1.5}{v} - \frac{1}{-30} = \frac{1.5 - 1}{-20}$
Simplifying, $\frac{1.5}{v} = -\frac{0.5}{20} - \frac{1}{30} = -\frac{1}{40} - \frac{1}{30} = -\frac{7}{120}$. Thus, $v = -\frac{180}{7}$ cm $\approx -25.71$ cm.

Key Concept: Dispersion Basics

c) Dispersion
[Solution Description]
Dispersion is the phenomenon where white light splits into its constituent colors (spectrum) when passing through a prism due to varying refractive indices for different wavelengths.

Key Concept: Refraction Formula for Spherical Surfaces

c) 30 cm
[Solution Description]
For a spherical surface, the refraction formula is:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Here, $n_1 = 1$ (air), $n_2 = \frac{3}{2}$, $u = \infty$ (parallel beam), and $R = 10$ cm (radius). Substituting these values:
$\frac{\frac{3}{2}}{v} - \frac{1}{\infty} = \frac{\frac{3}{2} - 1}{10}$
Simplifying:
$\frac{3}{2v} = \frac{1}{20} \implies v = 30 \text{ cm}$

Key Concept: Equivalent Focal Length

d) $-20 \, \text{cm}$
[Solution Description]
For two lenses in contact, the equivalent focal length $f$ is given by:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
For the given lenses: $f_1 = 20 \, \text{cm}$ (convex) and $f_2 = -10 \, \text{cm}$ (concave). Substituting these values:
$\frac{1}{f} = \frac{1}{20} + \frac{1}{-10} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20}$
Thus, the equivalent focal length is $f = -20 \, \text{cm}$.

Key Concept: Focal Length Formula

b) $-12$ cm
[Solution Description]
For a double concave lens, the focal length $f$ is given by:
$\frac{1}{f} = (n - 1)\left( \frac{1}{R_1} + \frac{1}{R_2} \right)$
Substituting $n = 1.5$, $R_1 = 10$ cm, and $R_2 = 15$ cm:
$\frac{1}{f} = (1.5 - 1)\left( \frac{1}{10} + \frac{1}{15} \right) = 0.5 \left( \frac{3 + 2}{30} \right) = 0.5 \times \frac{5}{30} = \frac{1}{12}$
Therefore, $f = -12$ cm (negative because it's a concave lens).

Key Concept: Reducing Aperture Size

b) It reduces the effect of marginal rays by limiting light to the paraxial region.
[Solution Description]
Stopping down the aperture reduces the size of the lens opening, allowing only paraxial rays (near the optical axis) to pass through. Since paraxial rays focus closer to the ideal focal point, this reduces the effect of marginal rays and thus minimizes spherical aberration.

Key Concept: Ray Diagram for Convex Lens

b) At twice the focal length on the opposite side of the lens
[Solution Description]
When an object is placed at twice the focal length ($u = 2f$) of a convex lens, we use the lens formula:
$\frac{1}{v} - \frac{1}{-2f} = \frac{1}{f}$
Simplifying, $\frac{1}{v} + \frac{1}{2f} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} - \frac{1}{2f} = \frac{1}{2f}$
Thus, $v = 2f$. The image will be formed at twice the focal length on the other side of the lens and will be real, inverted, and of the same size as the object.

Key Concept: Concave Lens, Magnification, Virtual Image

c) $2.0$
[Solution Description]
Given: $u = -15$ cm, $f = -10$ cm. Using lens formula:
$\frac{1}{v} - \frac{1}{-15} = \frac{1}{-10} \implies \frac{1}{v} = -\frac{1}{10} + \frac{1}{15} = -\frac{1}{30}$
Thus, $v = -30$ cm. Magnification $m = \frac{v}{u} = \frac{-30}{-15} = 2$.

Key Concept: Image Formation by Concave Lens

a) Virtual, erect, and 10 cm from the lens on the object side
[Solution Description]
Given: Focal length ($f$) = $-15$ cm (negative for concave lens), Object distance ($u$) = $-30$ cm (negative as per sign convention).
Using lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substitute the values:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-15} \implies \frac{1}{v} + \frac{1}{30} = -\frac{1}{15}$
Solve for $v$:
$\frac{1}{v} = -\frac{1}{15} - \frac{1}{30} = -\frac{2}{30} - \frac{1}{30} = -\frac{3}{30} = -\frac{1}{10}$
So, $v = -10$ cm. The negative sign indicates that the image is virtual and located 10 cm from the lens on the same side as the object.

Key Concept: Focal Length of Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that the focal length of a plano-convex lens with given parameters is 22.5 cm. Using the Lens Maker's Formula for a plano-convex lens:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R}\right)$
Substituting $n = 1.5$ and $R = 11.25\,\text{cm}$:
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{11.25}\right) = 0.5 \times \frac{1}{11.25} = \frac{1}{22.5}$
Thus, $f = 22.5\,\text{cm}$, confirming the Assertion is true.
The Reason correctly explains why the formula reduces to $\frac{1}{f} = (n - 1)\left(\frac{1}{R}\right)$ for a plano-convex lens (where one surface is flat, making $\frac{1}{\infty} = 0$). Therefore, both are true, and the Reason explains the Assertion.

Key Concept: Refraction at spherical surface

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Given values are $n_1 = 1$, $n_2 = 2$, $R = +10$ cm (convex surface), and $u = -20$ cm (real object).
Step 2: Substitute these into the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Step 3: Compute $\frac{2}{v} - \frac{1}{-20} = \frac{2 - 1}{10}$.
Step 4: Simplify to get $\frac{2}{v} + \frac{1}{20} = \frac{1}{10}$.
Step 5: Further simplify to $\frac{2}{v} = \frac{1}{10} - \frac{1}{20} = \frac{1}{20}$.
Step 6: Solve for $v$ as $v = \frac{2 \times 20}{1} = +40$ cm.
Step 7: Since $v$ is positive, the image is real (Assertion is true). The Reason correctly explains why a positive $v$ indicates a real image (Reason is true and correct explanation).

Key Concept: Lens Formula - Basic Application

c) 15 cm
[Solution Description]
Use the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Here, $u = -30$ cm (object distance is negative as per sign convention) and $f = +10$ cm (positive for convex lens).
Plugging in values: $\frac{1}{v} - \frac{1}{-30} = \frac{1}{10}$.
Solving gives $\frac{1}{v} + \frac{1}{30} = \frac{1}{10}$.
Subtract $\frac{1}{30}$ from both sides: $\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$.
Thus, $v = 15$ cm.

Key Concept: Refraction Formula

b) $-90$ cm
[Solution Description]
Using the refraction formula: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Substituting the given values: $\frac{1.5}{v} - \frac{1}{-15} = \frac{1.5 - 1}{10}$.
Simplifying: $\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{10}$.
Further simplifying: $\frac{1.5}{v} = 0.05 - 0.0667 = -0.0167$.
Solving for $v$: $v = \frac{1.5}{-0.0167} \approx -90$ cm.

Key Concept: Converging nature, Focal length

c) 26.67 cm
[Solution Description]
The lens maker's formula for a lens in air is:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Here, $\mu = 1.5$, $R_1 = +20$ cm (convex surface), and $R_2 = -40$ cm (concave surface). Plugging in the values:
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{-40}\right)$
$\frac{1}{f} = 0.5\left(\frac{1}{20} + \frac{1}{40}\right)$
$\frac{1}{f} = 0.5\left(\frac{2}{40} + \frac{1}{40}\right) = 0.5 \times \frac{3}{40} = \frac{1.5}{40}$
Therefore:
$f = \frac{40}{1.5} = 26.\overline{6} \text{ cm}$

Key Concept: Refraction from glass to air: n₂ = 1

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: Understand the assertion (A). It states that for a convex spherical surface separating glass ($n_1 = 1.5$) and air ($n_2 = 1$), the image formed when the object is placed in glass is always virtual.
Step 2: Recall the formula for refraction at a spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Step 3: For a convex surface, $R > 0$. Since $n_2 < n_1$ (air has a lower refractive index than glass), the right-hand side of the equation becomes negative.
Step 4: If the object is placed in glass, $u$ is negative (real object). Solving for $v$ with typical values shows it comes out negative, indicating a virtual image on the same side as the object. Thus, Assertion (A) is true.
Step 5: Examine Reason (R). It states that for a convex surface with $n_2 < n_1$, refracted rays diverge away from the principal axis. This is correct because the light bends away from the normal when passing from denser (glass) to rarer (air) medium, causing divergence.
Step 6: However, this divergence explains why the image is virtual (as refracted rays appear to come from a point behind the surface), making Reason (R) the correct explanation for Assertion (A).

Key Concept: Ray through focus, Image formation in convex lens

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a convex lens, when a ray passes through the first focus $F'$, it emerges parallel to the principal axis after refraction. This is because the first focus $F'$ is defined as the point where rays diverging from it (or appearing to converge towards it) become parallel to the principal axis upon refraction. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Key Formula

b) The magnification of the optical system.
[Solution Description]
The formula $m = \frac{v}{u}$ defines the magnification ($m$), where $v$ is the image distance and $u$ is the object distance. For minimal aberration, optimizing these distances helps reduce spherical aberration by ensuring proper ray convergence.

Key Concept: Magnification combination

c) $-2$
[Solution Description]
First, find the equivalent focal length ($F$) of the combination using:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2},$
where $f_1 = +10$ cm (convex) and $f_2 = -20$ cm (concave).
Substituting values:
$\frac{1}{F} = \frac{1}{10} + \frac{1}{-20} = \frac{2 - 1}{20} = \frac{1}{20}.$
Thus, $F = +20$ cm.
Now, use the lens formula to find the image distance ($v$):
$\frac{1}{v} - \frac{1}{u} = \frac{1}{F}, \quad u = -30 \text{ cm.}$
Substituting:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}.$
Simplifying:
$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}.$
Thus, $v = +60$ cm.
Magnification is:
$m = \frac{v}{u} = \frac{60}{-30} = -2.$

Key Concept: Refraction at a Convex Spherical Surface

d) 60 cm to the right of the pole
[Solution Description]
For a parallel beam, the object distance $u$ is at infinity. Using the formula $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$, with $u = \infty$, $n_1 = 1$, $n_2 = 1.5$, and $R = 20$ cm (positive for convex surface):
$\frac{1.5}{v} - \frac{1}{\infty} = \frac{1.5 - 1}{20}$
$\frac{1.5}{v} = 0 + \frac{0.5}{20}$
$\frac{1.5}{v} = 0.025$
$v = \frac{1.5}{0.025} = 60$ cm.
The positive sign indicates the image is real and formed 60 cm to the right of the pole.

Key Concept: Refraction from air to glass

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the speed of light decreases when it travels from air to glass, which is true because the refractive index ($n$) of glass is higher than air ($n_1 = 1$). According to the relation $v = \frac{c}{n}$, where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium, as $n$ increases, $v$ decreases. The reason correctly explains this phenomenon by stating that the refractive index of glass is greater than that of air. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Ray Diagrams for Concave Lens, Lens Formula

a) Virtual, erect, and at 6 cm from the lens
[Solution Description]
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where $u = -15$ cm (real object), $f = -10$ cm (concave lens). Substitute the values: $\frac{1}{v} - \frac{1}{-15} = \frac{1}{-10}$. Rearranging: $\frac{1}{v} + \frac{1}{15} = -\frac{1}{10}$. Now, $\frac{1}{v} = -\frac{1}{10} - \frac{1}{15}$. To solve this, find the common denominator (30): $\frac{1}{v} = -\frac{3}{30} - \frac{2}{30} = -\frac{5}{30} = -\frac{1}{6}$. Therefore, $v = -6$ cm. Since $v$ is negative, the image is virtual and forms on the same side as the object. The magnification ($m$) is $\frac{v}{u} = \frac{-6}{-15} = 0.4$, indicating the image is diminished and erect.

Key Concept: Thin Lens Formula

b) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
[Solution Description]
The thin lens formula relates the object distance ($u$), image distance ($v$), and focal length ($f$) of a lens. The correct formula is:
$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Key Concept: Combined Focal Length and Power, Cases of Lens Combinations

c) Assertion is true, but Reason is false.
[Solution Description]
First, calculate the combined focal length using the formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$. Substituting the values: $\frac{1}{f} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$. Thus, $f = -60$ cm, which means the combination behaves as a diverging lens (negative focal length). The assertion is true.
Now, check the reason: The combined power is indeed $P = P_1 + P_2$. However, the statement that the combination will **always** behave as a diverging lens if one lens is concave is false because it depends on the magnitudes of $f_1$ and $f_2$. For example, if $f_1 = 10$ cm (convex) and $f_2 = -15$ cm (concave), the combined focal length is positive ($f = 30$ cm), so the combination behaves as a converging lens. Thus, the reason is false.

Key Concept: Power of Lenses

c) 2 diopters
[Solution Description]
The power $P$ of a lens is given by $P = \frac{1}{f}$, where $f$ is the focal length in meters. Substituting $f = 0.5$ meters, we get:
$P = \frac{1}{0.5} = 2 \text{ diopters}$

Key Concept: Image Shift, Combined Lenses

a) $\frac{1}{60}$ D
[Solution Description]
For the first lens ($P_1 = 5$ D), the object distance $u$ can be found using the lens formula:
$\frac{1}{v} - \frac{1}{u} = P_1 \implies \frac{1}{30} - \frac{1}{u} = 5$
Solving for $u$ gives:
$-\frac{1}{u} = 5 - \frac{1}{30} = \frac{150 - 1}{30} = \frac{149}{30} \implies u = -\frac{30}{149} \, \text{m}$
For the combined lens ($P = P_1 + P_2$), the new image distance is $20$ cm:
$\frac{1}{20} - \frac{1}{u} = P_1 + P_2$
Substituting $u$ and $P_1$:
$\frac{1}{20} + \frac{149}{30} = 5 + P_2$
Simplifying:
$\frac{3 + 298}{60} = 5 + P_2 \implies \frac{301}{60} \approx 5.0167 = 5 + P_2$
Thus:
$P_2 \approx 0.0167 \, \text{D}$
However, this calculation seems complex; let's revisit the problem for a simpler solution. Alternatively, subtracting the two lens equations directly:
$\left(\frac{1}{20} - \frac{1}{30}\right) = P_2 \implies \frac{1}{60} = P_2$
Thus, $P_2 = \frac{1}{60}$ D.

Key Concept: Lens Maker's Formula, Magnification, Refractive Index

b) $-0.714$
[Solution Description]
Using Lens Maker's formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.6 - 1)\left(\frac{1}{20} - \frac{1}{\infty}\right) = 0.6 \times \frac{1}{20} = 0.03$
Thus, $f = \frac{100}{3}$ cm. Using lens formula for $u = -40$ cm:
$\frac{1}{v} - \frac{1}{-40} = \frac{3}{100} \implies v = 28.57 \, \text{cm}$
Magnification $m = \frac{v}{u} = \frac{28.57}{-40} = -0.714$.

Key Concept: Gaussian Lens Equation

c) $15$ cm
[Solution Description]
Using the Gaussian lens equation:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given $u = -30$ cm (object distance is negative by convention) and $f = 10$ cm:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{10}$
Simplifying:
$\frac{1}{v} + \frac{1}{30} = \frac{1}{10}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$
Thus, the image distance $v = 15$ cm.

Key Concept: Sign conventions and behavior of concave surfaces

b) Image is always virtual and erect
[Solution Description]
For a concave surface separating denser to rarer medium, the image formed is always virtual. This is because the light rays diverge after refraction, making the image appear on the same side as the object. Using the given parameters ($n_1 = 1.6$, $n_2 = 1.2$, $R = -10$ cm), we can confirm this behavior by plugging into the refraction formula. However, since it's a general property of concave surfaces separating denser to rarer media, the exact calculation isn't necessary here.

Key Concept: Lens Maker's Formula - Plano-Convex Lens

d) 20 cm
[Solution Description]
For a plano-convex lens, the lens maker’s formula simplifies to:
$\frac{1}{f} = (n - 1) \left( \frac{1}{R} \right)$
Given: $n = 1.5$, $R = 10$ cm. Substituting these values:
$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{10} \right) = 0.5 \times 0.1 = 0.05$
Taking reciprocal gives:
$f = \frac{1}{0.05} = 20 \text{ cm}$

Key Concept: Equal Focal Lengths

b) 15 cm
[Solution Description]
When the medium on both sides of a lens is the same, the first and second focal lengths are numerically equal, i.e., $f_1 = f_2$. Therefore, if $f_1 = 15$ cm, then $f_2 = 15$ cm as well.

Key Concept: Magnification Sign Convention

b) When the image is virtual and erect
[Solution Description]
For a convex lens, magnification ($m$) is positive when the image formed is virtual. A virtual image occurs when the object is placed within the focal length of the lens, making both object distance ($u$) and image distance ($v$) negative.

Key Concept: Aperture Effect

d) The exposure increases
[Solution Description]
Decreasing the f-number (e.g., from f/16 to f/4) increases the size of the aperture, allowing more light to reach the sensor. Since shutter speed is constant, this directly increases the exposure of the image. The correct answer is "The exposure increases."

Key Concept: Lens Maker's Formula

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the lens maker's formula is applicable to thin lenses with small apertures, which is true as per the syllabus. The reason explains that the formula assumes small angles of incidence and negligible lens thickness compared to radii of curvature, which is also true and correctly explains why the formula is valid for thin lenses. Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Combination of Lenses

c) 12 cm
[Solution Description]
The combined focal length ($f$) for two lenses in contact is given by:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Given: $f_1 = 20$ cm, $f_2 = 30$ cm.
Substituting the values:
$\frac{1}{f} = \frac{1}{20} + \frac{1}{30} = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12}$
Thus, $f = 12$ cm.

Key Concept: Total Internal Reflection

b) 38.68$^\circ$
[Solution Description]
The critical angle ($\theta_c$) is calculated using:
$\sin \theta_c = \frac{n_2}{n_1}$
Here, $n_1 = 1.6$ (denser medium), $n_2 = 1$ (air).
Substituting the values:
$\sin \theta_c = \frac{1}{1.6} = 0.625$
Taking inverse sine:
$\theta_c = \sin^{-1}(0.625) \approx 38.68^\circ$

Key Concept: Refraction at Spherical Surfaces, Lens Maker's Formula

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that for a biconvex lens with $R_1 = R_2 = R$ and $n = 1.5$, the focal length is $f = R$. Let us verify this using the lens maker's formula:
Given the lens maker's formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
For a biconvex lens, $R_1$ is positive and $R_2$ is negative (as per sign convention). Substituting $R_1 = R$ and $R_2 = -R$:
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{R} - \frac{1}{-R}\right) = 0.5 \left(\frac{2}{R}\right) = \frac{1}{R}$
Thus, $f = R$, which matches the assertion.
The reason states that the lens maker's formula reduces to $\frac{1}{f} = (n - 1)\left(\frac{2}{R}\right)$ for equal radii of curvature. This is true only if both radii are treated as positive (which is not correct due to the sign convention for a biconvex lens). Hence, the reason misrepresents the formula for this case.
Therefore, the assertion is true, but the reason is false.

Key Concept: Myopia and Hypermetropia Correction

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify the assertion and reason:
The power of the lens is given as $P = -4 \text{D}$. Using the formula $P = \frac{1}{f}$, we calculate the focal length:
$f = \frac{1}{-4} = -0.25 \text{m} = -25 \text{cm}.$
In myopia, the corrective lens forms a virtual image of distant objects at the far point of the eye. Thus, the lens's focal length should match the far point distance (25 cm). Here, both assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Combination of Two Thin Lenses in Contact

a) -20 cm
[Solution Description]
The equivalent focal length $f$ for two lenses in contact is given by:
$\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting $f_1 = 20 \, \text{cm}$ and $f_2 = -10 \, \text{cm}$:
$\frac{1}{f} = \frac{1}{20} + \frac{1}{-10} = \frac{1}{20} - \frac{1}{10} = -\frac{1}{20}$
Hence, $f = -20 \, \text{cm}$.

Key Concept: Concave Surface Refraction Formula

c) $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
[Solution Description]
The formula for refraction at a concave spherical surface is derived under small-angle approximation and is given by:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Here, $R$ is negative for a concave surface, and $v$ is always negative, indicating a virtual image.

Key Concept: Focal Length Sign Convention

b) Negative
[Solution Description]
For a concave lens, the principal focus is virtual and lies on the same side as the incident light. Hence, its focal length is negative according to the sign convention.

Key Concept: Lens Maker's Formula

c) 25 cm
[Solution Description]
Using the lens maker's formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Given: $n = 1.6$, $R_1 = 30$ cm, $R_2 = -30$ cm. Substituting these values:
$\frac{1}{f} = (1.6 - 1)\left(\frac{1}{30} - \frac{1}{-30}\right)$
Simplify:
$\frac{1}{f} = 0.6 \left(\frac{1}{30} + \frac{1}{30}\right) = 0.6 \times \frac{2}{30} = \frac{1.2}{30}$
$f = \frac{30}{1.2} = 25 \text{ cm}$
The focal length is 25 cm.

Key Concept: Ray Diagrams for Convex Lens, Image Formation Shifts

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when an object is placed between the first focus ($F$) and the optical centre of a convex lens, the image is virtual and enlarged. This is true because, in this position, the refracted rays diverge and appear to meet behind the lens, forming a virtual, erect, and larger image.
The reason explains that for a convex lens, if the object distance $u$ is less than the focal length $f$, the image distance $v$ is negative (virtual image). This is also correct, as per the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$. Here, $0 < u 0$, implying $v$ is negative (virtual image). Further, the reason correctly explains why the image is virtual, supporting the assertion. Thus, both are true, and the reason correctly explains the assertion.

Key Concept: Plano-convex lens, Apparent position

c) 22.5 cm below the lens
[Solution Description]
For refraction at the convex surface ($R = +15$ cm, $n_1 = 1$, $n_2 = 1.5$) and object distance $u = -10$ cm:
$\frac{1.5}{v} - \frac{1}{-10} = \frac{1.5 - 1}{15}$
Solving: $\frac{1.5}{v} + 0.1 = 0.0333$
$\frac{1.5}{v} = -0.0667$
$v = -22.5$ cm (virtual image, same side as object).

Key Concept: Magnification for Concave Lens

b) 0.6
[Solution Description]
For a concave lens, $f = -15 \text{ cm}$, $u = -10 \text{ cm}$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{-10} = \frac{1}{-15}$
Simplifying:
$\frac{1}{v} = -\frac{1}{15} - \frac{1}{10} = -\frac{5}{30} = -\frac{1}{6}$
Thus, $v = -6 \text{ cm}$.
Magnification $m$ is:
$m = \frac{v}{u} = \frac{-6}{-10} = 0.6$
The positive sign indicates a virtual and erect image.

Key Concept: Camera focus, depth of field

c) 40.8 m
[Solution Description]
Hyperfocal distance ($H$) is calculated as:
$H = \frac{f^2}{Nc} + f$
However since aperture ($N$) isn't given, we'll use the standard calculation ignoring second term:
$H \approx \frac{f^2}{c} = \frac{35^2}{0.03} = \frac{1225}{0.03} \approx 40833 \text{ mm} = 40.833 \text{ m}$
The closest option is approximately 40.8 meters.

Key Concept: Effect of Aperture

c) Decreases depth of field
[Solution Description]
A larger aperture (smaller f-number) decreases the depth of field, making the background more blurred, while a smaller aperture (higher f-number) increases it.

Key Concept: Lens Formula Application

c) 0.33
[Solution Description]
Given: $f = -6$ cm, $u = -12$ cm. Using lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values:
$\frac{1}{v} - \frac{1}{-12} = \frac{1}{-6} \implies \frac{1}{v} + \frac{1}{12} = -\frac{1}{6}$
Solving for $v$:
$\frac{1}{v} = -\frac{1}{6} - \frac{1}{12} = -\frac{2}{12} - \frac{1}{12} = -\frac{3}{12} = -\frac{1}{4}$
Thus, $v = -4$ cm. Magnification ($m$) is given by:
$m = \frac{v}{u} = \frac{-4}{-12} = \frac{1}{3}$
Since $|m| < 1$, the image is diminished.

Key Concept: Ray Diagrams for Concave Lens, Ray Tracing

b) -12 cm
[Solution Description]
For a concave lens, a ray parallel to the principal axis diverges after refraction and appears to come from the second focus ($F$). Here, the virtual focus is 12 cm from the lens, so the focal length $f$ is equal to the distance of this point from the lens. Since it's a concave lens, $f$ is negative. Thus, $f = -12$ cm.

Key Concept: Combined power of lenses

c) $+2D$
[Solution Description]
For two lenses in contact, the combined power is the algebraic sum of their individual powers:
$P = P_1 + P_2 = +5D + (-3D) = +2D$
The combination behaves like a converging lens since the resultant power is positive.

Key Concept: Spherical Aberration

c) Spherical aberration
[Solution Description] Spherical aberration happens because spherical lenses do not perfectly focus parallel rays to a single point. Rays passing through the edges focus at a different point than those near the center.

Key Concept: Image Distance Sign

b) Negative
[Solution Description]
A concave lens forms a virtual image on the same side as the object. According to the sign convention, distances measured in the direction opposite to the incident light (same side as the object) are negative. Thus, $v$ is negative.

Key Concept: Power of a Lens

c) 2 D
[Solution Description]
The power $P$ of a lens is given by $P = \frac{1}{f}$, where $f$ is the focal length in meters. Here, $f = 50\, \text{cm} = 0.5\, \text{m}$. So, $P = \frac{1}{0.5} = 2\, \text{D}$.

Key Concept: Ray through focus → emerges parallel to axis

a) Emerges parallel to the principal axis
[Solution Description]
For a concave lens, the first focus $F'$ is located 15 cm from the lens on the opposite side of the object. A ray directed towards the first focus of a concave lens behaves differently than in a convex lens.
According to the property of concave lenses, any ray appearing to go towards the first focus $F'$ will emerge parallel to the principal axis after refraction.
Here, since the ray is directed towards the first focus (virtual focus for the concave lens), it will refract and emerge parallel to the principal axis.

Key Concept: Lens Maker’s Formula, Radii of Curvature

a) $2.5 \, \text{cm}$
[Solution Description]
For an equiconvex lens, $R_1 = R$ and $R_2 = -R$. Using the Lens Maker’s Formula:
$$ \frac{1}{10} = (1.5 - 1)\left( \frac{1}{R} - \frac{1}{-R} \right) $$
Simplifying:
$$ \frac{1}{10} = 0.5 \times \frac{2}{R} \implies R = 10 \, \text{cm} $$
When one surface is silvered, the system behaves as a combination of a lens and a mirror with radius of curvature $R = 10$ cm (focal length of mirror $f_m = \frac{R}{2} = 5$ cm).
The equivalent focal length $f_{eq}$ is given by:
$$ \frac{1}{f_{eq}} = \frac{2}{f_l} + \frac{1}{f_m} $$
Here, $f_l = 10$ cm (unchanged), so:
$$ \frac{1}{f_{eq}} = \frac{2}{10} + \frac{1}{5} = 0.4 \, \text{cm}^{-1} $$
Thus, $f_{eq} = 2.5$ cm.

Key Concept: Lens Maker’s Formula

b) 10 cm
[Solution Description]
Using the Lens Maker's Formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Given: $n = 1.6$, $R_1 = 10$ cm, $R_2 = -15$ cm.
Substituting the values:
$\frac{1}{f} = (1.6 - 1)\left(\frac{1}{10} - \frac{1}{-15}\right)$
Simplifying inside the parentheses:
$\frac{1}{f} = 0.6\left(\frac{1}{10} + \frac{1}{15}\right)$
Finding common denominator (30):
$\frac{1}{f} = 0.6\left(\frac{3 + 2}{30}\right) = 0.6\left(\frac{5}{30}\right) = 0.1$
Therefore:
$f = \frac{1}{0.1} = 10 \text{ cm}$

Key Concept: Numerical Problem

b) $-25.71$ cm
[Solution Description]
Using the formula:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substitute the given values:
$\frac{1.5}{v} - \frac{1}{-30} = \frac{1.5 - 1}{-20}$
Simplify:
$\frac{1.5}{v} + \frac{1}{30} = -\frac{0.5}{20}$
$\frac{1.5}{v} = -\frac{1}{40} - \frac{1}{30} = -\frac{7}{120}$
Solve for $v$:
$v = \frac{1.5 \times 120}{-7} = -25.71 \text{ cm}$
Since $v$ is negative, the image is virtual.

Key Concept: Magnification, Virtual image formation

b) -12 cm
[Solution Description]
Given: $f = -12$ cm (concave lens), magnification $m = +\frac{1}{2}$ (virtual image is upright and diminished).
Magnification formula: $m = \frac{v}{u}$
Thus, $v = \frac{u}{2}$.
Using thin lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substitute $v = \frac{u}{2}$:
$\frac{2}{u} - \frac{1}{u} = -\frac{1}{12}$
Simplify:
$\frac{1}{u} = -\frac{1}{12}$
So, $u = -12$ cm. The negative sign indicates the object is on the same side as the incident light.

Key Concept: Basic Optics

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the effective focal length of a microscope depends on the focal lengths of the objective and eyepiece lenses, which is true because the combined effect determines magnification. However, the reason incorrectly links magnification directly to the ratio of focal lengths—the correct formula involves the tube length and angular magnification. Thus, the assertion is true, but the reason is false.

Key Concept: Ray Diagram Concepts

c) A ray directed towards the optical center passes undeviated through the lens
[Solution Description]
According to the key points mentioned in the syllabus:
1. The refracted rays diverge at the focus after passing through the concave lens.
2. The image formed is always virtual, erect, and diminished.
3. The image lies between the lens and the focus.
Among the given options, only one matches these properties.

Key Concept: Aperture and Light Gathering Power

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that doubling the aperture diameter increases the light gathering power, which is correct because the light gathering power ($L$) is given by $L \propto D^2$, where $D$ is the aperture diameter. If $D$ is doubled, $L$ becomes $L \propto (2D)^2 = 4D^2$, meaning it quadruples. The reason correctly explains the assertion as it describes the proportionality relationship between light gathering power and aperture diameter squared.

Key Concept: Coordinate Geometry Sign Convention

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to the coordinate geometry sign convention:
1. The focal length of a converging lens is positive, while that of a diverging lens (concave lens) is negative. So Assertion (A) is true.
2. For a concave lens, the rays incident parallel to the principal axis appear to diverge from the principal focus after refraction. This principal focus is virtual and lies on the same side as the object, confirming Reason (R) is also true.
3. Moreover, Reason (R) correctly explains why the focal length is negative for a concave lens under this sign convention. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Lens Maker's Formula, Different Medium

b) 50 cm
[Solution Description]
The effective refractive index becomes $n_{\text{eff}} = \frac{n_{\text{lens}}}{n_{\text{liquid}}} = \frac{1.75}{1.4} = 1.25$.
Using the Lens Maker's Formula:
$\frac{1}{f} = (n_{\text{eff}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Here, $R_1 = +25$ cm (convex) and $R_2 = -25$ cm (convex from the other side).
Substitute:
$\frac{1}{f} = (1.25 - 1) \left( \frac{1}{25} - \frac{1}{-25} \right) = 0.25 \left( \frac{2}{25} \right) = \frac{0.5}{25} = \frac{1}{50}$
Thus, $f = 50$ cm in the liquid.

Key Concept: Refraction at a concave spherical surface

b) $-33.6$ cm to the left of the surface
[Solution Description]
Given: $n_1 = 1.33$, $n_2 = 1.5$, $R = -15$ cm (since it is concave), $u = -40$ cm (object distance is negative as per sign convention). Using the formula for refraction at a concave surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting the values:
$\frac{1.5}{v} - \frac{1.33}{-40} = \frac{1.5 - 1.33}{-15}$
Simplifying:
$\frac{1.5}{v} + \frac{1.33}{40} = \frac{0.17}{-15}$
Subtract $\frac{1.33}{40}$ from both sides:
$\frac{1.5}{v} = -\frac{0.17}{15} - \frac{1.33}{40}$
Calculate $-\frac{0.17}{15} - \frac{1.33}{40} = -\frac{6.8}{600} - \frac{19.95}{600} = -\frac{26.75}{600}$
Thus:
$\frac{1.5}{v} = -\frac{26.75}{600}$
Solve for $v$:
$v = \frac{1.5 \times 600}{-26.75} \approx -33.64 \, \text{cm}$
The image is formed approximately 33.64 cm to the left of the surface (virtual image).

Key Concept: Image Formation by Convex Surface

c) $\infty$
[Solution Description]
Using the formula for refraction at a convex surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting $n_1 = 1$, $n_2 = 1.5$, $u = -20$ cm, and $R = 10$ cm,
$\frac{1.5}{v} - \frac{1}{-20} = \frac{1.5 - 1}{10}$
Simplifying,
$\frac{1.5}{v} + 0.05 = 0.05$
Solving for $v$,
$\frac{1.5}{v} = 0 \\ \\ \Rightarrow v = \infty$
Thus, the image is formed at infinity.

Key Concept: Combination of Convex Lens and Plane Mirror, Virtual Image Formation

b) $u = f$
[Solution Description]
When the image coincides with the object in this setup, the rays must be falling normally on the plane mirror after refraction through the lens. This means the object must be at the focal point of the lens because only then will the refracted rays become parallel and reflect back along the same path. Therefore, $u = f$.
Step-by-step explanation:
1) The plane mirror reflects light back along the incident path.
2) For the image to coincide with the object, the rays must hit the mirror normally.
3) Only light coming from the focal point emerges parallel after refraction through the convex lens.
4) Hence, the object must be placed at the focal point ($u = f$).

Key Concept: Lens Maker’s Formula

b) 10 cm
[Solution Description]
Given: $n = 1.6$, $R_1 = 10$ cm, $R_2 = -15$ cm (since the second surface is concave from the incoming light's perspective). Applying the Lens Maker's formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Substituting the values:
$\frac{1}{f} = (1.6 - 1)\left(\frac{1}{10} - \frac{1}{-15}\right) = 0.6 \left(\frac{1}{10} + \frac{1}{15}\right) = 0.6 \left(\frac{3 + 2}{30}\right) = \frac{0.6 \times 5}{30} = \frac{3}{30} = \frac{1}{10}$
Thus, $f = 10$ cm.

Key Concept: Combination of two refracting surfaces, Equivalent focal length

d) Assertion is false, but Reason is true.
[Solution Description]
For two thin convex lenses in contact, the equivalent focal length formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$ gives a positive value since both $f_1$ and $f_2$ are positive. However, whether the combination forms a real image depends on the object distance relative to the combined focal length. If the object is placed within the focal length of the combination ($u < f$), a virtual image is formed. Thus, the Assertion is false. The Reason is true as the formula correctly represents the equivalent focal length for convex lenses. Therefore, the correct answer is d).

Key Concept: Magnification by Combination of Lenses

c) 6
[Solution Description]
The total magnification ($m$) for multiple lenses in combination is the product of individual magnifications: $m = m_1 \times m_2$. Substituting the given values, $m = 2 \times 3 = 6$.

Key Concept: Image Formation by Concave Lens

b) Virtual and erect
[Solution Description]
A concave lens diverges the incident rays and always forms a virtual, erect, and diminished image for any position of the object. The image is formed between the lens and the focus on the same side as the object.

Key Concept: Chromatic aberration, Spherical aberration

b) The focal length decreases to $19.62\,\text{cm}$
[Solution Description]
Chromatic aberration occurs because different wavelengths of light have different refractive indices in the lens material. The focal length of a lens is given by the lensmaker's formula:
$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
For yellow light, $\mu = 1.52$ and $f = 20\,\text{cm}$. For blue light, $\mu_b = 1.53$. The ratio of focal lengths is inversely proportional to the difference in refractive indices:
$\frac{f_b}{f_y} = \frac{\mu_y - 1}{\mu_b - 1} = \frac{1.52 - 1}{1.53 - 1} = \frac{0.52}{0.53} \approx 0.9811$
Therefore, the focal length for blue light is $f_b = f_y \times 0.9811 = 20 \times 0.9811 \approx 19.62\,\text{cm}$.

Key Concept: Equivalent focal length, Power calculation

c) $-5D$
[Solution Description]
Let $P_1 = +4D$ (convex lens), $P_{comb} = -1D$. For lenses in contact, $P_{comb} = P_1 + P_2$.
Step 1: Substitute known values:
$-1D = +4D + P_2$
Step 2: Solve for $P_2$:
$P_2 = -1D - 4D = -5D$
The concave lens must have a power of $-5D$ to produce this combination.

Key Concept: Correction of Aberrations

c) Applying aperture stops or parabolic surfaces.
[Solution Description]
Spherical aberration can be reduced by using parabolic surfaces or aperture stops. Aperture stops limit the rays passing through the lens, reducing the effect of edge rays that cause blurring.

Key Concept: Microscope and Telescope Magnification

d) Assertion is false, but Reason is true.
[Solution Description]
For the microscope, the magnification formula is $M = \frac{v}{u} \times \frac{D}{f_e}$. While decreasing $f_e$ increases magnification, there is a practical limit due to lens aberrations and manufacturing constraints, making the assertion false. For the telescope, the angular magnification formula is $M = \frac{f_o}{f_e}$, showing that magnification is indeed inversely proportional to $f_e$, making the reason true.

Key Concept: Combination of Lenses, Power of Lenses

c) 15 D
[Solution Description]
The combined power $P$ of two lenses in contact is the sum of their individual powers:
$P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2}$
Substituting $f_1 = 10$ cm and $f_2 = 20$ cm:
$P = \frac{1}{0.1} + \frac{1}{0.2} = 10 + 5 = 15 \, \text{D}$

Key Concept: Combined Power of Two Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The power of a lens is defined as $P = \frac{1}{f}$. When two lenses are placed in contact, their powers add up because the deviation caused by one lens is followed by the other. Thus, for two lenses with powers $P_1$ and $P_2$, the combined power is indeed $P = P_1 + P_2$. Also, the reason correctly states that power is inversely proportional to focal length ($P \propto \frac{1}{f}$). This relationship explains why the assertion is true. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Conceptual Understanding

b) Higher magnification
[Solution Description]
A shorter focal length in the objective lens results in higher magnification. This is because the magnifying power is inversely proportional to the focal length—smaller focal lengths bend light more sharply, increasing image enlargement.

Key Concept: Cause of Dispersion

b) Because they have different frequencies
[Solution Description]
Different colors (wavelengths) of light travel at different speeds in a medium, causing them to refract at different angles. This is due to the dependence of refractive index on wavelength.

Key Concept: Refraction Formula for Spherical Surfaces

b) Image is formed at infinity
[Solution Description]
Using the refraction formula for spherical surfaces:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Given: $n_1 = 1$, $n_2 = 1.5$, $u = -50$ cm (object distance is negative as per sign convention), $R = 25$ cm. Substituting these values:
$\frac{1.5}{v} - \frac{1}{-50} = \frac{1.5 - 1}{25}$
Simplify and solve for $v$:
$\frac{1.5}{v} + \frac{1}{50} = \frac{0.5}{25}$
$\frac{1.5}{v} = \frac{0.5}{25} - \frac{1}{50}$
$\frac{1.5}{v} = \frac{1}{50} - \frac{1}{50} = 0$
This leads to no solution, meaning the image is formed at infinity.

Key Concept: Lens combination, Power of lenses

a) Converging lens with 1.67 D
[Solution Description]
The combined focal length ($F$) is given by:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$.
Substituting the values:
$\frac{1}{F} = \frac{1}{20} + \frac{1}{-30} = \frac{3 - 2}{60} = \frac{1}{60}$.
Thus, $F = 60$ cm.
The power of the combination is $P = \frac{1}{F} = \frac{1}{0.6} = 1.67$ D (since $F$ is in meters).
Since $F$ is positive, the combination behaves as a converging lens.

Key Concept: Combination of Lenses

b) 7.5 cm
[Solution Description]
When two identical lenses are placed in contact, the combined focal length $f'$ is given by:
$\frac{1}{f'} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}$
Given: $f = 15$ cm. Substituting the value:
$\frac{1}{f'} = \frac{2}{15}$
Taking reciprocal gives:
$f' = \frac{15}{2} = 7.5 \text{ cm}$

Key Concept: Refraction Formula Application

b) $-75$ cm
[Solution Description]
Using the refraction formula:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Substituting the given values:
$\frac{1.6}{-50} - \frac{1.2}{u} = \frac{1.6 - 1.2}{-25}$
Simplifying:
$-\frac{1.6}{50} - \frac{1.2}{u} = -\frac{0.4}{25}$
Multiply through by $-1$:
$\frac{1.6}{50} + \frac{1.2}{u} = \frac{0.4}{25}$
Simplify $\frac{0.4}{25}$ to $\frac{0.8}{50}$:
$\frac{1.6}{50} + \frac{1.2}{u} = \frac{0.8}{50}$
Subtract $\frac{1.6}{50}$ from both sides:
$\frac{1.2}{u} = \frac{0.8}{50} - \frac{1.6}{50} = -\frac{0.8}{50}$
Solving for $u$:
$u = \frac{1.2}{-0.8/50} = -\frac{1.2 \times 50}{0.8} = -75 \text{ cm}$
The object distance is $-75$ cm.

Key Concept: Camera Focus

b) 51.3 mm
[Solution Description]
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Given $f = 50$ mm $= 0.05$ m and $u = -2$ m (since object distance is taken as negative in lens formula conventions). Substituting the values:
$\frac{1}{v} - \frac{1}{-2} = \frac{1}{0.05}$
Simplifying:
$\frac{1}{v} + 0.5 = 20$
$\frac{1}{v} = 19.5$
$v = \frac{1}{19.5} \approx 0.0513 \text{ m or } 51.3 \text{ mm}$

Key Concept: Refraction at Concave Spherical Surface

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that for a concave spherical surface, the image formed is always virtual. This is true because as per the given derivation, the value of $v$ is always negative for all negative values of $u$, indicating the image is formed in the first medium and is virtual.
The Reason states that a concave surface curves away from the incident light, causing diverging rays which appear to come from a virtual image. This is also true and correctly explains why the image is virtual—the geometry of the concave surface ensures refraction causes divergence.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Lens Formula

b) $-10$ cm
[Solution Description]
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting $u = -30$ cm and $f = -15$ cm:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-15}$
Simplifying:
$\frac{1}{v} + \frac{1}{30} = -\frac{1}{15}$
$\frac{1}{v} = -\frac{1}{15} - \frac{1}{30} = -\frac{3}{30} = -\frac{1}{10}$
$v = -10 \text{ cm}$

Key Concept: Converging and Diverging Actions of Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A convex lens is thicker in the middle and thinner at the edges. It converges parallel rays of light to a point called the focal point. The reason states that in a convex lens, the base of each small-angled prism component is towards the principal axis, which causes the rays to bend towards it. This bending results in convergence. Here, both the assertion and reason are true, and the reason correctly explains why a convex lens converges light.

Key Concept: Hypermetropia and Lens Power

b) Convex lens
[Solution Description]
Hypermetropia is corrected using a convex (converging) lens with positive power. Since the power is $+4$ D, the lens must be convex.

Key Concept: Distances measured from pole

c) $\frac{n_2}{v'} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
[Solution Description]
The correct relation for refraction at a spherical surface from rarer medium $n_1$ to denser medium $n_2$ is given by:
$\frac{n_2}{v'} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
This formula describes how the distances are measured from the pole $P$ of the spherical surface.

Key Concept: Symmetry

d) Infinite
[Solution Description]
Due to complete spherical symmetry, any diameter can be a principal axis, meaning there are infinite principal axes that are all equivalent.

Key Concept: Focal Length Definition

b) 10 cm
[Solution Description]
The second focus (principal focus) is where parallel rays converge after passing through the lens. The distance from the optical center to this focus is the focal length ($f$). Here, $f = 10$ cm.

Key Concept: Power calculation, image formation, lens formula

a) $12 \text{ cm}$
[Solution Description]
First, find the focal length of the diverging lens:
$P = -5D$, so $f = \frac{1}{P} = \frac{1}{-5} = -0.2 \text{ m} = -20 \text{ cm}$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
Given $u = -30 \text{ cm}$ (object distance), $f = -20 \text{ cm}$:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-20}$.
Simplify:
$\frac{1}{v} + \frac{1}{30} = -\frac{1}{20}$.
Rearrange:
$\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = -\frac{3}{60} - \frac{2}{60} = -\frac{5}{60} = -\frac{1}{12}$.
Thus, $v = -12 \text{ cm}$.
The virtual image is formed at 12 cm from the lens on the same side as the object.
The correct answer is $12 \text{ cm}$.

Key Concept: Combined Power of Two Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The combined power of two lenses in contact is indeed given by $P = P_1 + P_2$. If $P_1$ is positive (convex lens) and $P_2$ is negative (concave lens), the net power depends on their magnitudes. Given that $|P_1| > |P_2|$, $P$ will be positive, meaning the combination behaves as a convex lens. Therefore, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Combination of Thin Lenses in Contact

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For two thin lenses in contact, the equivalent focal length $f$ is calculated using the formula $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$. The power $P$ of a lens is given by $P = \frac{1}{f}$, where $f$ is in meters. Therefore, the combined power is indeed the sum of the individual powers. The assertion is true, and the reason correctly explains why the assertion holds.

Key Concept: Telescope, Angular Magnification, Focal Length

c) 24
[Solution Description]
For normal adjustment, angular magnification ($M_{\infty}$) is:
$M_{\infty} = \frac{f_o}{f_e} = \frac{100}{5} = 20$
When the image is at the near point, the effective magnification increases. The new angular magnification ($M_D$) is:
$M_D = M_{\infty} \left(1 + \frac{f_e}{D}\right)$
Here, $D = 25 \text{ cm}$:
$M_D = 20 \left(1 + \frac{5}{25}\right) = 20 \times 1.2 = 24$

Key Concept: Magnification by a Convex Lens

b) -2
[Solution Description]
Given: Object distance ($u$) = -30 cm, Focal length ($f$) = +20 cm (for convex lens).
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting values: $\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$
Solving for $v$: $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$
Thus, $v = 60$ cm (real image as $v$ is positive).
Magnification ($m$) = $\frac{v}{u} = \frac{60}{-30} = -2$.

Key Concept: Image Formation by Concave Lens

c) Virtual, erect, 6.67 cm on the same side as the object
[Solution Description]
Using the Gaussian Lens Formula for a concave lens ($f = -10 \text{ cm}$) and object distance $u = -20 \text{ cm}$:
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-10} \implies \frac{1}{v} + \frac{1}{20} = -\frac{1}{10}$
Solving for $v$:
$\frac{1}{v} = -\frac{1}{10} - \frac{1}{20} = -\frac{2 + 1}{20} = -\frac{3}{20}$
Thus, $v = -\frac{20}{3} \text{ cm} \approx -6.67 \text{ cm}$. The image is virtual, erect, and formed on the same side as the object.

Key Concept: Chromatic Aberration

c) Chromatic aberration
[Solution Description] Chromatic aberration occurs because the refractive index of a lens varies with wavelength (color). This causes different colors to bend by different amounts and focus at different points.

Key Concept: Coma, Astigmatism

c) Coma
[Solution Description]
Coma is an off-axis aberration that occurs when light rays pass through different parts of the lens at different angles, causing the image of a point source to appear as a comet-shaped blur. Unlike spherical aberration (which affects on-axis rays) or astigmatism (which causes elongation of the image), coma specifically distorts the shape asymmetrically.

Key Concept: Sign Conventions, Linear Magnification

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that a convex lens forms a virtual image when the object is placed between the optical centre and the focus. This is true because for a convex lens, if the object is placed within the focal length ($u < f$), the image formed is virtual, erect, and magnified.
The Reason states that for a convex lens, the magnification for a virtual image is positive because the image distance $v$ is negative according to the sign convention. This is also correct because, for a virtual image formed by a convex lens, $v$ is negative (as per the sign convention where virtual images are on the same side as the object). The magnification formula is $m = \frac{v}{u}$, and since both $v$ and $u$ are negative (for virtual image and real object), the magnification turns out to be positive.
However, the Reason does not correctly explain the Assertion. The Assertion describes the condition for forming a virtual image, while the Reason explains why the magnification is positive. They are independent statements, with the Reason not being the explanation for the Assertion.

Key Concept: Basic Concept of Optical Instruments

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
A microscope indeed uses two convex lenses—the objective lens and the eyepiece. The objective lens forms a real and inverted intermediate image, which is further magnified by the eyepiece. Therefore, both Assertion (A) and Reason (R) are true, and Reason correctly explains why the microscope works.

Key Concept: Refraction at Concave Spherical Surface

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that a concave spherical surface forms a virtual image for any position of the object placed in the rarer medium, which is true as per the given syllabus. The reason (R) states that the refracted rays always diverge away from each other after refraction, which is also true. However, the divergence of refracted rays does not explain why the image formed is virtual. The correct explanation would involve the specific refractive index conditions and the geometry of the concave surface. Hence, both (A) and (R) are true, but (R) is NOT the correct explanation of (A).

Key Concept: Refraction at spherical surfaces, Lens maker's formula

c) 20 cm
[Solution Description]
For a silvered plano-convex lens, the net power is given by $P_{net} = 2P_L + P_M$ where $P_L$ is the lens power and $P_M$ is the mirror power.
First calculate lens power: Using lens maker's formula $\frac{1}{f_L} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Here $R_1 = 20$ cm (convex), $R_2 = \infty$ (plane). So $\frac{1}{f_L} = (1.5-1)\left(\frac{1}{20} - 0\right) = 0.025$ cm$^{-1}$.
Mirror power: $P_M = -\frac{1}{f_M} = -\frac{2}{R}$. Since plane surface is silvered, $R = \infty$, so $P_M = 0$.
Net power: $P_{net} = 2 \times 0.025 + 0 = 0.05$ cm$^{-1}$.
Effective focal length: $f_{net} = \frac{1}{0.05} = 20$ cm.

Key Concept: Thin lens formula

b) $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
[Solution Description]
The thin lens formula relates the object distance ($u$), image distance ($v$), and focal length ($f$) as follows:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
This equation is fundamental in determining any of these parameters when the other two are known.

Key Concept: Refraction through curved spherical surfaces

d) Assertion is false, but Reason is true.
[Solution Description]
To verify the assertion, we use the refraction formula for a spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Given:
- $n_1 = 1$ (air), $n_2 = 1.5$ (glass)
- $u = -20$ cm (object distance, negative as per sign convention)
- Let $R = +30$ cm (radius of curvature for convex surface, positive as per sign convention).
Substituting the values:
$\frac{1.5}{v} - \frac{1}{-20} = \frac{1.5 - 1}{30}$
Simplifying:
$\frac{1.5}{v} + 0.05 = \frac{0.5}{30} \approx 0.0167$
Further simplification gives:
$\frac{1.5}{v} = 0.0167 - 0.05 = -0.0333$
Solving for $v$:
$v = \frac{1.5}{-0.0333} \approx -45 \text{ cm}$
However, the assertion claims $v = +60$ cm, which is incorrect based on this calculation. The reason provides the correct formula but the assertion's conclusion is false. Thus, the reason is true, but the assertion is false.

Key Concept: Refraction Formula for Spherical Surfaces

d) 60 cm in glass
[Solution Description]
Using the refraction formula for a spherical surface:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
Here, $u = \infty$ (parallel beam), so $\frac{n_1}{u} = 0$.
Substituting the values:
$\frac{1.5}{v} = \frac{1.5 - 1}{20} \Rightarrow \frac{1.5}{v} = \frac{0.5}{20} \Rightarrow v = \frac{1.5 \times 20}{0.5} = 60 \, \text{cm}$
The image forms at 60 cm in the second medium (glass).

Key Concept: Lens Formula and Magnification

b) $-2$
[Solution Description]
Given: Object distance $u = -30 \text{ cm}$, Focal length $f = 20 \text{ cm}$.
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting values:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20}$
Simplifying:
$\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$
Thus, $v = 60 \text{ cm}$.
Magnification $m$ is given by:
$m = \frac{v}{u} = \frac{60}{-30} = -2$
The negative sign indicates that the image is inverted.

Key Concept: Thin Lens Formula, Magnification

a) 0.33
[Solution Description]
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, where $u = -30$ cm (object distance is negative as per sign convention) and $f = -15$ cm (diverging lens has negative focal length). Solving for $v$:
$\frac{1}{v} - \frac{1}{-30} = \frac{1}{-15}$
Simplifying gives:
$v = -10 \, \text{cm}$
Magnification is calculated as $m = \frac{v}{u} = \frac{-10}{-30} = \frac{1}{3}$.

Key Concept: Power calculation, corrective lenses

a) $-2D$
[Solution Description]
To correct myopia, a diverging lens is used to shift the far point to infinity.
Given far point $u = -50 \text{ cm} = -0.5 \text{ m}$ and desired image distance $v = -\infty$.
Using the lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-\infty} - \frac{1}{-0.5} = 0 + \frac{1}{0.5} = 2$.
Power of the lens:
$P = \frac{1}{f} = 2D$.
However, since it's a diverging lens (as per myopia correction), the power is negative:
$P = -2D$.
The correct answer is $-2D$.

Key Concept: Refraction at Convex Surface

d) $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
[Solution Description]
The formula for refraction at a convex spherical surface is:
$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$
where $n_1$ is the refractive index of the rarer medium, $n_2$ is the refractive index of the denser medium, $u$ is the object distance (negative as per sign convention), $v$ is the image distance, and $R$ is the radius of curvature (positive for convex surface).

Key Concept: Spherical Aberration

c) It blurs the image due to non-uniform convergence of light rays.
[Solution Description]
Spherical aberration occurs because rays passing through the edges of a spherical lens converge at a different point compared to those passing near the center. This results in a blurred image as not all rays meet at a single focal point.

Key Concept: Resolving Power of Telescope

b) It improves by a factor of 2
[Solution Description]
The resolving power of a telescope is given by $\theta = \frac{1.22 \lambda}{D}$, where $\theta$ is the smallest angular separation that can be resolved. If $D$ is doubled, $\theta$ becomes half because $\theta$ is inversely proportional to $D$. Therefore, the resolving power improves by a factor of 2.

Key Concept: Lens Formula - Sign Convention

b) Positive
[Solution Description]
According to the sign convention, the focal length ($f$) of a convex lens is positive because it forms real images and has a converging effect on light.

Key Concept: Lens Formula - Concave Lens

a) -10 cm
[Solution Description]
For a concave lens, the focal length ($f$) is negative ($f = -15$ cm). The object distance ($u$) is $-30$ cm.
Using the lens formula: $\frac{1}{v} - \frac{1}{-30} = \frac{1}{-15}$.
Simplifying: $\frac{1}{v} + \frac{1}{30} = -\frac{1}{15}$.
Further simplification gives $\frac{1}{v} = -\frac{1}{15} - \frac{1}{30} = -\frac{2 + 1}{30} = -\frac{3}{30} = -\frac{1}{10}$.
Thus, $v = -10$ cm (negative sign indicates virtual image).

Key Concept: Parabolic Reflectors

c) They focus parallel rays to a single focal point
[Solution Description]
Parabolic reflectors focus incoming parallel rays of light to a single focal point, minimizing spherical aberrations and improving image quality. This property makes them ideal for applications requiring precise focusing, such as telescopes.

Key Concept: Ray Diagrams for Concave Lens, Magnification

b) -6.67 cm
[Solution Description]
Given magnification $m = 0.25$ and object distance $u = -20$ cm. Magnification for a concave lens is given by $m = \frac{v}{u}$. So, $0.25 = \frac{v}{-20}$, hence $v = -5$ cm (virtual image). Now, using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$: $\frac{1}{-5} - \frac{1}{-20} = \frac{1}{f}$. Simplifying: $-\frac{1}{5} + \frac{1}{20} = \frac{1}{f}$. Find the common denominator (20): $-\frac{4}{20} + \frac{1}{20} = -\frac{3}{20} = \frac{1}{f}$. Thus, $f = -\frac{20}{3} \approx -6.67$ cm. The negative sign indicates a concave lens.

Key Concept: Lateral Displacement

b) Decreases and becomes negligible for a thin lens
[Solution Description]
For a thick lens, rays passing through the optical centre experience lateral displacement. As the thickness decreases, this displacement reduces and becomes negligible for a thin lens, where the ray passes straight without deviation.

Key Concept: Basic Concept

a) Combination of objective and eyepiece lenses
[Solution Description]
The effective focal length of a compound microscope depends mainly on the combination of its objective lens and eyepiece lenses. The magnification power is directly influenced by the focal lengths of these lenses. Shorter focal lengths result in higher magnification.

Key Concept: Magnification and Power of a Lens Combination

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The power of the convex lens is calculated as $P_1 = \frac{1}{f_1} = \frac{1}{0.20} = +5$ D. The power of the concave lens is $P_2 = \frac{1}{f_2} = \frac{1}{-0.10} = -10$ D. The total power of the combination is $P = P_1 + P_2 = 5 - 10 = -5$ D. Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Ray Diagrams for Convex Lens

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
When an object is placed beyond $2F$ of a convex lens, two key rays are considered:
1. A ray parallel to the principal axis refracts and passes through the second focus $F'$.
2. A ray passing through the optical center continues undeviated.
These two rays intersect on the other side of the lens, forming a real, inverted, and diminished image between $F'$ and $2F'$. Thus, both Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Power calculation, unit conversion

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
To verify Assertion (A):
Given power ($P$) = $-5 \text{ D}$. Using the formula for power:
$P = \frac{1}{f \text{ (in metres)}}$
Substituting $P$:
$-5 = \frac{1}{f}$
Solving for $f$:
$f = \frac{1}{-5} = -0.2 \text{ m}$
Thus, the focal length calculated is $-0.2 \text{ m}$, which matches the given value in Assertion (A).
For Reason (R):
The negative power indeed indicates a diverging lens (concave), making Reason (R) true.
Since both Assertion (A) and Reason (R) are true, and Reason correctly explains the assertion, the correct option is:

Key Concept: Lens Maker's Formula, Image Formation by Convex Lens

a) Real, inverted, at 18.18 cm on the opposite side of the lens
[Solution Description]
First, calculate the focal length $f$ using the Lens Maker's Formula:
$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
Here, $n = 1.6$, $R_1 = 20$ cm (since it's plano-convex, $R_2 = \infty$):
$\frac{1}{f} = (1.6 - 1) \left( \frac{1}{20} \right) = 0.03 \, \text{cm}^{-1}$
So, $f = 33.\overline{3}$ cm.
Next, use the Gaussian form of the lens equation to find the image distance $v$:
$\frac{1}{v} - \frac{1}{-40} = \frac{1}{33.\overline{3}}$
Solving for $v$:
$\frac{1}{v} = \frac{1}{33.\overline{3}} + \frac{1}{40} \approx 0.055 \, \text{cm}^{-1}$
$v \approx 18.18$ cm (positive, so real image).
The magnification $m = -\frac{v}{u} = -\frac{18.18}{40} \approx -0.4545$ (inverted and diminished).

Key Concept: Focal Length Formula

b) $-2R$
[Solution Description]
For a plano-concave lens, one surface is plane ($R_1 = \infty$) and the other is concave ($R_2 = -R$). Using the lensmaker's formula:
$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
Substituting values:
$\frac{1}{f} = (1.5 - 1)\left(0 - \frac{1}{-R}\right) = 0.5 \times \frac{1}{R}$
Therefore, $f = -2R$.

Key Concept: Power of Combined Lenses

b) 2 D
[Solution Description]
For lenses in contact, the net power ($P$) is the algebraic sum of individual powers: $P = P_1 + P_2$. Given $P_1 = +5$ D and $P_2 = -3$ D:
$P = 5 + (-3) = 2 \, \text{D}$