Class 12 Physics (Part 2) Chapter 19 Wave Nature of Light : Huygens’ Principle

Key Concept: Illustration of Direction of Propagation

b) The angle of reflection is equal to the angle of incidence.
[Solution Description]
When a plane wavefront is incident on a plane reflector, the direction of the reflected wavefront follows the law of reflection: the angle of incidence equals the angle of reflection. Both angles are measured with respect to the normal to the reflecting surface.

Key Concept: Wavefront and Direction of Propagation

c) Rays are normal to the wavefront.
[Solution Description]
According to the syllabus, in a homogeneous isotropic medium, the rays (direction of propagation) are always normal (perpendicular) to the wavefront. This is a fundamental property of wave propagation in such media.

Key Concept: Speed of Light in Different Media

b) $2.26 \times 10^8 \, \text{m/s}$
[Solution Description]
The speed of light in a medium is given by:
$v = \frac{c}{n}$
where $c = 3 \times 10^8 \, \text{m/s}$ and $n = 1.33$. Substituting the values:
$v = \frac{3 \times 10^8}{1.33} \approx 2.26 \times 10^8 \, \text{m/s}$
Therefore, the speed of light in water is approximately $2.26 \times 10^8 \, \text{m/s}$.

Key Concept: Newton’s Corpuscular Theory

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 is greater in denser media according to Newton’s corpuscular theory, which is true based on the key points provided in the syllabus. The reason explains that this increase in speed is due to the attraction of corpuscles by the denser medium, which aligns with Newton’s explanation for refraction. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Reflection of Waves

d) The angle of incidence equals the angle of reflection
[Solution Description]
The law of reflection states that the angle of incidence (\$i\$) is equal to the angle of reflection (\$r\$), i.e., \$i = r\$. Hence, the correct option must state this equality.

Key Concept: Rays and Wavefronts, Huygens' Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that in a homogeneous isotropic medium, the direction of energy propagation (rays) is always perpendicular to the wavefront. This is true because, in such media, the wavefront expands uniformly, and the rays (energy propagation direction) are always normal to the wavefront. The reason explains Huygens’ principle, which describes how wavefronts propagate by considering each point on a wavefront as a source of secondary wavelets. Since these secondary wavelets combine to form the new wavefront, the rays remain perpendicular to the wavefront. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Snell's Law and Wavefront Derivation

a) $\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$
[Solution Description]
During the derivation of Snell's Law, it is shown that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two media. From the geometry of the incident and refracted wavefronts, we derive:
$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$
This is the fundamental relationship in Snell's Law, where $v_1$ is the speed of light in the first medium and $v_2$ is the speed in the second medium.

Key Concept: Refractive Index and Speed of Light

b) 1.25
[Solution Description]
The refractive index of medium A relative to medium B is given by:
$n = \frac{v_B}{v_A}$
where $v_A = 2 \times 10^8 \, \text{m/s}$ and $v_B = 2.5 \times 10^8 \, \text{m/s}$. Substituting the values:
$n = \frac{2.5 \times 10^8}{2 \times 10^8} = 1.25$
Therefore, the refractive index of medium A relative to medium B is 1.25.

Key Concept: Law of Reflection

d) $v t$
[Solution Description]
The radius of the secondary wavelet is given by the distance covered by light in time $t$ with velocity $v$. According to Huygens' principle, the secondary wavelet from point A grows to a radius equal to $AB' = v t$. Therefore, the correct answer is $v t$.

Key Concept: Huygens’ Principle, Reflection, Refraction

d) 133.22°
[Solution Description]
Using Snell's law for refraction:
$n_1 \sin i = n_2 \sin r$
$1.0 \times \sin 30^\circ = \sqrt{3} \times \sin r$
$\sin r = \frac{0.5}{\sqrt{3}} \approx 0.2887$
$r \approx \sin^{-1}(0.2887) \approx 16.78^\circ$
The angle of reflection equals the angle of incidence, so the reflected ray makes 30° with the normal. The angle between the reflected and refracted rays is:
$180^\circ - 30^\circ - 16.78^\circ = 133.22^\circ$

Key Concept: Energy Transmission

d) $I \propto \frac{1}{r^2}$
[Solution Description]
For a spherical wavefront, the energy spreads uniformly over the surface area of a sphere, which increases as $4\pi r^2$. Thus, intensity ($I$), being power per unit area, follows the inverse square law: $I \propto \frac{1}{r^2}$.

Key Concept: No Explanation for Polarization

b) Because polarization is a property of waves
[Solution Description]
Polarization involves the orientation of light vibrations in a specific plane, which is a property of waves. Since the corpuscular theory treated light as particles, it could not account for this wave-like behavior, leading to its limitation.

Key Concept: Energy Transmission in Spherical Wavefront

b) $A \propto \frac{1}{r}$
[Solution Description]
For a spherical wavefront, the energy spreads out over the surface area of an expanding sphere. The intensity ($I$) decreases as $I \propto \frac{1}{r^2}$. Since intensity is proportional to the square of the amplitude ($I \propto A^2$), the amplitude decreases as $A \propto \frac{1}{r}$.

Key Concept: Huygens' Principle, Wavelet Distance

b) $2 \times 10^{-10}$ s
[Solution Description]
According to Huygens' principle, the wavelet from point A travels outward in all directions after reflection. The distance covered by the wavelet is given as 6 cm.
Time taken is calculated as:
$t = \frac{d}{v} = \frac{6 \times 10^{-2}}{3 \times 10^8} = 2 \times 10^{-10} \text{ s}$

Key Concept: Behavior of Wavefronts in Refraction

d) Spherical converging
[Solution Description]
A convex lens converges parallel rays of light to a single point called the focus. According to Huygens’ principle, each point on the incident plane wavefront acts as a source of secondary spherical wavelets. After refraction through the convex lens, these wavelets converge to form a spherical wavefront that converges towards the focus. Hence, the refracted wavefront is spherical and converging.

Key Concept: Direction of Propagation

c) Rays are perpendicular to the wavefront.
[Solution Description]
In a homogeneous isotropic medium, rays represent the direction of energy propagation and are always perpendicular (normal) to the wavefront. This is because the disturbance propagates in the direction normal to the wavefront in such media.

Key Concept: Wave Theory

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that Huygens' wave theory explains the rectilinear propagation of light using secondary wavelets, which is true. The reason given is also correct because Huygens' principle indeed describes how every point on a wavefront emits secondary wavelets that spread out with the same speed as the original wave. Furthermore, the reason correctly explains the assertion because the superposition of these secondary wavelets leads to the formation of a new wavefront, demonstrating rectilinear propagation. Thus, both are true, and the reason is the correct explanation of the assertion.

Key Concept: Refraction of Light – Huygens’ Explanation

b) $\frac{\lambda}{n}$
[Solution Description]
The wavelength of light in a medium ($\lambda_m$) decreases compared to its wavelength in vacuum ($\lambda$) due to the refractive index ($n$) of the medium. It is given by the relation $\lambda_m = \frac{\lambda}{n}$.

Key Concept: Wavefront Derivation Step

c) $BA' = v_1 t$
[Solution Description]
The distance covered by the wavefront in time $t$ while traveling at speed $v_1$ is given by the formula:
$\text{Distance} = \text{Speed} \times \text{Time}$
Substituting the given values:
$BA' = v_1 t$

Key Concept: Plane wavefront, Cylindrical wavefront

b) Cylindrical wavefront has diverging rays, plane wavefront has parallel rays
[Solution Description]
For a cylindrical wavefront, the rays are perpendicular to the cylindrical surface and diverge from the line source. In contrast, for a plane wavefront, the rays are parallel straight lines perpendicular to the wavefront. Therefore, the key difference is that rays from a cylindrical wavefront diverge radially outward from the line source, while rays from a plane wavefront do not diverge; they remain parallel.

Key Concept: Forward wavefront, Huygens' Principle

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that the forward wavefront is the only observable wavefront because the secondary wavelets have maximum intensity in this direction. This is true as per Huygens’ principle, which explains that the forward wavefront is formed by the constructive interference of secondary wavelets.
The reason claims that the backward wavefront cancels out due to destructive interference. While it is true that the backward wavefront is not observed, Huygens’ principle does not directly state cancellation due to destructive interference; rather, it suggests the intensity varies from maximum in the forward direction to zero in the backward direction. Thus, the reason partially explains the assertion but introduces an incorrect detail about destructive interference leading to cancellation.
Therefore, the Assertion is true, but the Reason is false.

Key Concept: Spherical Wavefront, Huygens’ Principle

b) Spherical wavefront
[Solution Description]
According to Huygens’ principle, every point on the wavefront acts as a source of secondary wavelets. For a small aperture ($d << R$), the curvature of the wavefront can be neglected locally, and the secondary wavelets will behave as if originating from a point source. Thus, the wavefront beyond the aperture will be spherical.

Key Concept: Speed of Light in Medium, Refractive Index

c) $1.5$
[Solution Description]
The refractive index ($n$) of a medium is defined as $n = \frac{c}{v}$, where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium. Given $v = 2.0 \times 10^8 \, \text{m/s}$ and $c = 3.0 \times 10^8 \, \text{m/s}$,
$n = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5$
Therefore, the refractive index of the glass is $1.5$.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to wave optics, the frequency (\$\nu\$) of light is determined by the source and does not change when light travels through different media. However, the speed of light (\$v\$) and wavelength (\$\lambda\$) change in a medium due to the refractive index (\$n\$), as \$v = \frac{c}{n}\$ and \$\lambda' = \frac{\lambda}{n}\$. The Reason correctly states that the energy of photons depends solely on frequency (\$E = h\nu\$), which remains constant. Thus, the Assertion and Reason are both true, and the Reason correctly explains the Assertion.

Key Concept: Effect of Refraction on Wavelength

Correct Answer option (either of a), b), c) or d)) Option text
[Solution Description]
Solution: According to the syllabus, when light passes from one medium to another, its frequency remains unchanged, but its speed and wavelength change. The speed of light in a medium is given by $v = \frac{c}{n}$, where $c$ is the speed of light in vacuum and $n$ is the refractive index. Since $n > 1$, the speed decreases in the denser medium. The wavelength in the medium is given by $\lambda' = \frac{\lambda}{n}$, which means the wavelength also decreases proportionally. Thus, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Key Concept: Wavefront transformation, lenses

b) Plane wavefront moving backward
[Solution Description]
When a plane wavefront is incident normally on a concave mirror, it gets converted to a spherical converging wavefront heading toward the focal point. At the focal plane, all secondary wavelets arrive simultaneously because they travel equal optical paths. The wavefront at the focal plane becomes planar again but with reversed curvature - diverging outward from the focal point.
Working:
1. Initial plane wave becomes spherical converging
2. After reaching focus, it transforms to spherical diverging
3. On focal plane, it appears as a plane wave again but propagating outward

Key Concept: Wavefront and Huygens' Principle

d) Wavefront
[Solution Description]
The surface where all particles of the medium are in the same phase of oscillation is defined as a wavefront. According to the given concept, this is the correct definition.

Key Concept: Wavefront and Huygens' Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a spherical wavefront is produced by a point source of light. According to Huygens' principle, a point source emits waves in all directions, and at any instant, all particles of the medium lying on a sphere centered at the source oscillate in the same phase, forming a spherical wavefront. Thus, the assertion is true.
The reason states that in a homogeneous isotropic medium, all points on the spherical wavefront are at the same phase. This is correct because the disturbance from the point source reaches all points on the spherical wavefront simultaneously, meaning they oscillate in phase. Hence, the reason is also true and correctly explains the assertion.

Key Concept: Effect of Refraction on Wavelength

b) The wavelength decreases
[Solution Description]
When light passes from air to glass, the refractive index of glass is greater than 1. According to the relation $\lambda_w = \frac{\lambda}{n}$, since $n > 1$, the wavelength in glass ($\lambda_w$) decreases compared to its wavelength in air ($\lambda$).

Key Concept: Huygens’ Principle and Forward Wavefront

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Huygens’ principle, every point on a wavefront acts as a source of secondary wavelets. These wavelets spread out in all directions with the speed of light in the medium. The forward wavefront is formed by the envelope of these secondary wavelets in the forward direction, which determines the new wavefront. This explains why the forward wavefront forms the observed light path. The reason states that the intensity of the secondary wavelets varies from maximum in the forward direction to zero in the backward direction, which explains why the backward wavefront is not observed. Both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Amplitude and Intensity

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the intensity ($I$) of a light wave is directly proportional to the square of its amplitude ($A$). This is correct as per the syllabus, where $I \propto A^2$. The reason states that the energy carried by a light wave depends on its amplitude, which is also true because the amplitude determines the energy of the wave. Moreover, the reason correctly explains the assertion since the dependence of intensity on the square of amplitude arises from the fact that energy is determined by amplitude. Hence, both are true, and the reason is the correct explanation of the assertion.

Key Concept: Comparison of Theories

b) Interference
[Solution Description]
Newton's corpuscular theory could explain reflection and rectilinear propagation but failed to explain interference, diffraction, and polarization.

Key Concept: Refractive index and speed

b) $1.33$
[Solution Description]
The refractive index ($n$) is given by:
$n = \frac{c}{v}$
where $v$ is the speed of light in the medium. Substituting $v = 2.25 \times 10^8 \, \text{m/s}$ and $c = 3.0 \times 10^8 \, \text{m/s}$:
$n = \frac{3.0 \times 10^8}{2.25 \times 10^8} = 1.33$

Key Concept: Behavior of Wavefronts - Plane Wavefront Reflection

a) Plane
[Solution Description]
When a plane wavefront strikes a plane reflecting surface, the angle of incidence is equal to the angle of reflection as per the laws of reflection. The reflected wavefront remains plane because the surface is flat and does not alter the planar nature of the incoming wavefront.

Key Concept: Wavefront and Huygens’ Principle

c) Plane
[Solution Description]
Using Huygens' principle, each point on the incident wavefront acts as a source of secondary wavelets. The reflected wavefront is the envelope of these secondary wavelets in the forward direction. Since the angle of incidence equals the angle of reflection ($i = r$), the shape of the reflected wavefront remains plane.

Key Concept: Wavefront Types

d) Intensity decreases inversely with the square of distance
[Solution Description]
For a spherical wavefront, the intensity $I$ is inversely proportional to the square of the distance $r$ from the source: $I \propto \frac{1}{r^2}$. This is because the energy spreads uniformly over the surface area of an expanding sphere.

Key Concept: Refraction, Wavelength Change

b) $\frac{n^2}{\lambda}$
[Solution Description]
Optical path length = $nD$ where $D$ is physical path length.
Reduced wavelength in medium: $\lambda_m = \frac{\lambda}{n}$.
Therefore, ratio = $\frac{nD}{\lambda/n} = \frac{n^2 D}{\lambda}$.
For unit path length ($D = 1$), ratio = $\frac{n^2}{\lambda}$.

Key Concept: Wavelength in denser medium

b) $\frac{\lambda}{2}$
[Solution Description]
The wavelength of light in a denser medium is given by:
$\lambda' = \frac{\lambda}{n_2}$
Given $n_2 = 2$, we substitute the value:
$\lambda' = \frac{\lambda}{2}$
Therefore, the wavelength in the second medium is half of its original wavelength.

Key Concept: Intensity of a wave

b) $I = \frac{1}{2} \epsilon_0 c E_0^2$
[Solution Description]
The intensity $I$ of an electromagnetic wave is given by the formula:
$I = \frac{1}{2} \epsilon_0 c E_0^2$
where $\epsilon_0$ is the permittivity of free space, $c$ is the speed of light, and $E_0$ is the amplitude of the electric field.

Key Concept: Optical Path

b) $nD$
[Solution Description]
The optical path is defined as the product of the refractive index and the physical distance traveled by the light in the medium. Therefore, the optical path is given by $nD$.

Key Concept: Rays are normals to wavefronts

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
In a homogeneous isotropic medium, light travels in straight lines called rays. These rays represent the direction of energy propagation and are always perpendicular (normal) to the wavefront. A plane wavefront has parallel rays, while a spherical wavefront has radially diverging or converging rays. Thus, both the Assertion (A) and Reason (R) are true, and Reason correctly explains Assertion.

Key Concept: Nature of Light in Corpuscular Theory

c) Corpuscles (particles)
[Solution Description]
The syllabus states that Newton proposed light consists of tiny particles called corpuscles. This was a fundamental assumption of his theory.

Key Concept: Comparing speeds of different colors in glass

c) Blue light
[Solution Description]
The speed of light in glass depends on the refractive index as:
$v = \frac{c}{n}$
Lower the refractive index, higher the speed. Since blue light has a higher refractive index ($n_b = 1.665$) compared to red light ($n_r = 1.645$), it will travel slower in glass.

Key Concept: Effect of Refraction on Wavelength

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to the syllabus, when light passes from one medium to another, its frequency remains unchanged but its wavelength changes. The wavelength in a medium ($\lambda'$) is given by $\lambda' = \frac{\lambda}{n}$, where $\lambda$ is the wavelength in vacuum and $n$ is the refractive index of the medium. Since $n > 1$ for water compared to air, the wavelength decreases in water. The speed of light in a medium is also related to the refractive index by $v = \frac{c}{n}$. Thus, both Assertion (A) and Reason (R) are true. However, Reason (R) explains why the speed changes, not directly why the wavelength decreases. Therefore, Reason (R) is not the correct explanation of Assertion (A).

Key Concept: Huygens’ Principle, Spherical Wavefront Construction

a) Secondary wavelets do not emit energy backward.
[Solution Description]
Huygens' principle states that every point on a wavefront acts as a secondary source emitting wavelets. The new wavefront is the envelope of these wavelets. Due to the directional nature of secondary wavelets, their intensity is maximum in the forward direction and zero in the backward direction. Hence, no backward wavefront is observed.

Key Concept: Laws of Reflection, Angle Calculation

d) 120°
[Solution Description]
The angle of incidence ($i$) is the angle between the incident wavefront and the normal, which is 60°. According to the law of reflection, the angle of reflection ($r$) is also 60°.
The angle between the incident wavefront and the reflected wavefront is the sum of the angles they make with the reflecting surface.
Since each wavefront makes an angle of 90° - 60° = 30° with the surface, the total angle between them is:
$\theta = 30^\circ + 30^\circ = 60^\circ$

Key Concept: Frequency, speed of light, wavelength

b) $6 \times 10^{14} \, \text{Hz}$
[Solution Description]
The frequency of light remains unchanged when passing from one medium to another. Therefore, the frequency in water will be the same as in air, which is $6 \times 10^{14} \, \text{Hz}$.

Key Concept: Wavefront Types

c) Spherical wavefront
[Solution Description]
A point source emits waves uniformly in all directions. Therefore, at any instant, all points equidistant from the source lie on a sphere, forming a spherical wavefront.

Key Concept: Refraction and Snell's Law

b) $\sin i / \sin r = v_1 / v_2$
[Solution Description]
Snell's law relates the angle of incidence ($i$) and the angle of refraction ($r$) to the refractive indices of the two media:
$\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{21}$
where $n_{21}$ is the refractive index of the second medium relative to the first.

Key Concept: Refraction, Speed of Light

b) $1.875 \times 10^8 \, \text{m/s}$
[Solution Description]
The speed of light in a medium ($v$) is given by:
$v = \frac{c}{n}$
where $c$ is the speed of light in vacuum and $n$ is the refractive index. Substituting the given values:
$v = \frac{3 \times 10^8 \, \text{m/s}}{1.6} = 1.875 \times 10^8 \, \text{m/s}$

Key Concept: Wavefronts, Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
For a spherical wavefront, the amplitude ($A$) decreases as $\frac{1}{r}$, and the intensity ($I$) decreases as $\frac{1}{r^2}$ due to the spreading of energy over a larger area. The Assertion correctly states that the amplitude decreases with distance, and the Reason provides the correct explanation for this phenomenon by mentioning the inverse square law relationship of intensity with distance, which indirectly explains the amplitude decrease since $I \propto A^2$. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Wave Theory

b) Luminiferous ether
[Solution Description]
Huygens proposed that light travels as waves in a hypothetical medium called 'luminiferous ether.'

Key Concept: Newton’s Corpuscular Theory, Limitations, Comparison with Huygens’ Wave Theory

c) Assertion is true, but Reason is false.
[Solution Description]
The Assertion states that Newton’s corpuscular theory predicts the speed of light to be greater in denser media, which is true as per the syllabus. The Reason claims that this prediction arises because corpuscles experience an attractive force in denser media, thereby increasing their velocity. However, this explanation is incorrect. Newton's theory assumes that forces act on corpuscles **perpendicular** to the surface, not necessarily increasing their speed but bending their path. The increase in speed in denser media was an inherent assumption of the theory, not explicitly explained by attractive forces. Hence, the Assertion is true, but the Reason is false.

Key Concept: Huygens' Principle, Wavefront Construction

c) The new wavefront is a sphere of radius $\frac{v_2}{v_1} BA'$ centered at A.
[Solution Description]
According to Huygens' principle, every point on the wavefront AB acts as a secondary source. For refraction:
1. The time taken for point B to reach point A' (on the boundary) is $\frac{BA'}{v_1}$.
2. During this time, the wavelet from point A propagates into medium 2 with speed $v_2$, forming a spherical wavelet of radius $v_2 t = v_2 \left(\frac{BA'}{v_1}\right) = \frac{v_2}{v_1} BA'$.
3. The new wavefront in medium 2 is the tangent drawn to all such secondary wavelets.
Therefore, the secondary wavelet from point A contributes a spherical wavelet of radius $\frac{v_2}{v_1} BA'$.

Key Concept: Huygens' Principle, Wave Propagation

b) Inversely proportional to distance
[Solution Description]
For a spherical wavefront, the energy spreads over the surface area of an expanding sphere ($4\pi r^2$). Since intensity $I$ is power per unit area and amplitude $A$ is proportional to $\sqrt{I}$, the amplitude decreases as the inverse of distance from the source: $A \propto \frac{1}{r}$.

Key Concept: Wavefront Types

c) Plane
[Solution Description]
When a point source emits light, the wavefront near the source is spherical due to the radial propagation of waves. However, at a large distance from the source, the radius of curvature becomes very large, making the wavefront appear almost plane. Therefore, the wavefront will be plane.

Key Concept: Wavefront properties

c) Homogeneous isotropic medium
[Solution Description]
According to the syllabus, this property holds specifically for homogeneous isotropic media, where the wavefront propagates uniformly in all directions.

Key Concept: Wavelength in a medium

b) $376 \, \text{nm}$
[Solution Description]
The wavelength in the medium ($\lambda_m$) is related to the wavelength in air ($\lambda$) and the refractive index ($n$) as:
$\lambda_m = \frac{\lambda}{n}$
Substituting $\lambda = 500 \, \text{nm}$ and $n = 1.33$:
$\lambda_m = \frac{500}{1.33} \approx 376 \, \text{nm}$

Key Concept: Plane Wavefront

b) Plane
[Solution Description]
When the light source is very far away (like a distant star), the curvature of the wavefront becomes negligible, and the wavefront appears flat or planar. Hence, the wavefront is plane.

Key Concept: Huygens' Principle

b) The wavefront remains plane
[Solution Description]
According to Huygens' principle, each point on the wavefront acts as a source of secondary wavelets. After reflection from a plane surface, the envelope of these secondary wavelets forms a new plane wavefront.

Key Concept: Wavefront and Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a plane wavefront incident on a plane mirror reflects as a plane wavefront, which is correct because the reflection preserves the shape of the wavefront (only its direction changes). The reason correctly explains the assertion by invoking Huygens’ principle, where secondary wavelets reconstruct the reflected plane wavefront. Hence, both are true, and the reason explains the assertion.

Key Concept: Wavefront expansion and Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a spherical wavefront expands outward with time, increasing its radius uniformly. This is true because as per Huygens’ principle, each point on the wavefront acts as a new source of secondary wavelets, which propagate spherically outward at the same speed in a homogeneous isotropic medium. Consequently, the envelope of these secondary wavelets forms a new spherical wavefront with a uniformly increased radius.
The reason correctly explains the assertion because it highlights that Huygens’ principle accounts for the uniform expansion by ensuring every secondary wavelet contributes equally to the new wavefront. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Relative Refractive Index

c) 1.5
[Solution Description]
The refractive index of glass with respect to air is calculated using the formula:
$\mu = \frac{\text{Speed of light in air}}{\text{Speed of light in glass}} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$
So, the correct answer is 1.5.

Key Concept: Types of Wavefronts

c) Spherical wavefront
[Solution Description]
A spherical wavefront is generated when waves originate from a point source, as they propagate uniformly in all directions. Thus, the correct option is spherical wavefront.

Key Concept: Refraction

c) Due to increase in speed
[Solution Description]
Newton proposed that corpuscles experience an attractive force near the surface of a denser medium, increasing their velocity and causing bending towards the normal.

Key Concept: Wavefront

c) Spherical
[Solution Description]
A point source emits waves uniformly in all directions. The wavefront at any instant is a sphere centered at the point source because all points equidistant from the source lie on a spherical surface.

Key Concept: Refraction of Light, Wavelength Calculation

c) $400 \, \text{nm}$
[Solution Description]
The wavelength of light in a medium ($\lambda'$) is given by $\lambda' = \frac{\lambda}{n}$, where $\lambda$ is the wavelength in vacuum and $n$ is the refractive index of the medium. Here, $\lambda = 600 \, \text{nm}$ and $n = 1.5$.
$\lambda' = \frac{600}{1.5} = 400 \, \text{nm}$
So, the wavelength in the medium is $400 \, \text{nm}$.

Key Concept: Behavior of Wavefronts

a) Plane
[Solution Description]
According to Huygens' principle and the given syllabus, when a plane wavefront strikes a plane reflecting surface, the reflected wavefront remains plane. This satisfies both the first and second laws of reflection.

Key Concept: Laws of Refraction Using Huygens’ Principle

b) 1.5
[Solution Description]
According to Huygens' principle, when light travels from one medium to another, the ratio of sines of the angle of incidence ($i$) and angle of refraction ($r$) is equal to the refractive index of the second medium relative to the first ($n_{21}$). Mathematically:
$\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n_{21} = \frac{n_2}{n_1}$
Here, $n_1 = 1$ (air) and $n_2 = 1.5$ (glass), so:
$\frac{\sin i}{\sin r} = 1.5$

Key Concept: Frequency in Refraction

c) Frequency
[Solution Description]
The frequency of light remains unchanged when it passes from one medium to another. According to wave optics, only the speed and wavelength change while the frequency stays constant.

Key Concept: Huygens' Principle and Reflection of Light

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Step 1: According to Huygens' principle, every point on a wavefront acts as a source of secondary wavelets. These wavelets spread out in all directions with the same speed as the original wave.
Step 2: For reflection, consider a plane wavefront incident on a reflecting surface. The time taken for the edge of the wavefront to reach the surface is calculated, and during this time, the secondary wavelet from the point of incidence spreads out.
Step 3: The tangent to this secondary wavelet forms the reflected wavefront. From the geometry of the wavefronts, it can be shown that $\sin i = \sin r$, which implies $i = r$, where $i$ is the angle of incidence and $r$ is the angle of reflection.
Step 4: The reason correctly explains how the secondary wavelets lead to the formation of the reflected wavefront, which satisfies the law of reflection. Hence, both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Rays are perpendicular to the wavefront, Spherical wavefront divergence

c) The rays are radial and perpendicular to the spherical wavefront.
[Solution Description]
In a homogeneous isotropic medium, the rays are always perpendicular to the wavefront. For a spherical wavefront originating from a point source, the rays are radial lines emanating from the source. As the wavefront expands outward, the rays remain normal to the spherical surface at every point.

Key Concept: Huygens' Principle and Reflection

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets. When a plane wavefront strikes a plane reflecting surface, the secondary wavelets reconstruct the reflected wavefront. By analyzing the geometry of the wavefronts and using the time taken for the wavefront to travel, we derive $\sin i = \sin r$ (where $i$ is the angle of incidence and $r$ is the angle of reflection), leading to $i = r$. Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Wavefront properties

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Huygens' wave theory, a spherical wavefront is indeed produced by a point source of light because light travels uniformly in all directions from such a source, forming a sphere where all points vibrate in phase. The reason correctly explains the assertion as it states that all points on the wavefront are equidistant from the point source, which is why they form a spherical shape.
Therefore, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Wavefront

b) Wavefront
[Solution Description]
A wavefront is defined as a surface where all particles of the medium vibrate in the same phase. The shape of the wavefront depends on the nature of the source and the medium.

Key Concept: Proof of Laws of Reflection Using Huygens’ Principle

c) $i = r$
[Solution Description]
Huygens' principle demonstrates that for reflection:
$\sin i = \sin r$
This leads to the conclusion that the angle of incidence is equal to the angle of reflection ($i = r$). Additionally, the incident ray, reflected ray, and normal all lie in the same plane.

Key Concept: Spherical Wavefront Propagation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
According to Huygens’ principle, a spherical wavefront is formed when light diverges from a point source, as stated in the Assertion. The Reason correctly states that the intensity ($I$) of light follows an inverse-square law, $I \propto \frac{1}{r^2}$, where $r$ is the distance from the source. However, the Reason does not explain why the wavefront is spherical; it merely describes the variation of intensity with distance. Therefore, both Assertion and Reason are true, but the Reason is not the correct explanation of the Assertion.

Key Concept: Reflection of Light Waves, Time Calculations

b) $5 \times 10^{-10}$ s
[Solution Description]
The incident wavefront AB makes a 45° angle with the reflector XY. The distance from A to XY is 10 cm, so the perpendicular distance is $AD = 10$ cm. The distance BC along the reflector is 5 cm.
First, find the distance AC using the right triangle ABC:
$AC = \sqrt{AB^2 + BC^2}$
But AB = AD / sin(45°) = $10 / (1/\sqrt{2}) = 10\sqrt{2}$ cm.
Now, calculate AC:
$AC = \sqrt{(10\sqrt{2})^2 + 5^2} = \sqrt{200 + 25} = \sqrt{225} = 15 \text{ cm}$
Time taken:
$t = \frac{AC}{c} = \frac{15 \times 10^{-2}}{3 \times 10^8} = 5 \times 10^{-10} \text{ s}$

Key Concept: Wavefront and Rays

c) The reflected wavefront is plane, and the rays remain parallel but change direction.
[Solution Description]
According to Huygens' principle, after reflection from a plane surface, the wavefront remains plane. The rays, which are perpendicular to the wavefront, change direction such that the angle of incidence equals the angle of reflection ($i = r$). Thus, the reflected wavefront is plane, and the rays remain parallel but change direction.

Key Concept: Assertion-Reason

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Solution: Huygens’ principle states that every point on a wavefront is a source of secondary wavelets, which spread out in the forward direction at the same speed as the original wave. The assertion correctly describes this phenomenon. The reason also correctly states that these secondary wavelets travel with the same speed as the original wave. Moreover, the reason correctly explains why the assertion is true because the uniform propagation speed of secondary wavelets is fundamental to reconstructing the new wavefront. Thus, both assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Ray Direction and Wavefront

b) Perpendicular to the wavefront
[Solution Description]
In a homogeneous isotropic medium, rays represent the direction of propagation of the wave. They are always perpendicular (normal) to the wavefront at every point. For a spherical wavefront, the rays are radial lines originating from the point source.

Key Concept: Wavefront and Rays

c) The angle of reflection is 45\textdegree.
[Solution Description]
According to the laws of reflection:
1. The angle of incidence ($i$) equals the angle of reflection ($r$): $i = r$.
2. The incident ray, reflected ray, and normal lie in the same plane.
Given $i = 45\textdegree$, then $r = 45\textdegree$.
The reflected wavefront will also be a plane, and the rays will be perpendicular to it, making an angle of 45\textdegree with the normal.

Key Concept: Plane Wavefront Characteristics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion (A) states that in a plane wavefront, all particles oscillate in the same phase. This is true because a plane wavefront is defined as a surface where all particles are in the same phase of oscillation.
The reason (R) states that a plane wavefront is perpendicular to the direction of propagation. This is also true and directly explains why all particles on a plane wavefront oscillate in the same phase: since the wavefront is perpendicular to the propagation direction, any point on it is equidistant from the source, leading to identical phases.
Thus, both Assertion and Reason are true, and Reason correctly explains the Assertion.

Key Concept: Refractive index

b) 1.5
[Solution Description]
The refractive index ($n$) is calculated using the formula:
$n = \frac{\text{speed of light in vacuum}}{\text{speed of light in medium}}$
Substituting the given values:
$n = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$

Key Concept: Snell's Law

c) Both speed and wavelength decrease
[Solution Description]
When light enters a denser medium (water), its speed decreases since $v = \frac{c}{n}$, where $n > 1$. The frequency remains constant but wavelength decreases proportionally since $\lambda_w = \frac{\lambda_a}{n}$.

Key Concept: Huygens' principle, Wavefront expansion

b) $v \alpha t$ radians
[Solution Description]
The question involves non-uniform propagation of secondary wavelets due to velocity variation.
Step 1: Consider two points on AB - point A ($x=0$) and point B ($x=L$). The velocity at A is $v_A = v(1 + \alpha \times 0) = v$, and at B is $v_B = v(1 + \alpha L)$.
Step 2: After time $t$, wavelet from A expands to radius $r_A = v t$ and from B to $r_B = v(1 + \alpha L)t$.
Step 3: The difference in radii creates an angle $\theta$ between old and new wavefronts: $\tan\theta = \frac{r_B - r_A}{L} = \frac{v(1 + \alpha L)t - v t}{L} = v \alpha t$
For small angles, $\theta \approx \tan\theta = v \alpha t$

Key Concept: Refraction: Frequency remains constant

c) Frequency
[Solution Description]
The frequency of light does not change when it moves from one medium to another. Only the speed and wavelength of light change based on the refractive index of the medium. Thus, the frequency remains constant.

Key Concept: Rays and Wavefronts

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that in a homogeneous isotropic medium, rays are always normal to the wavefront. This is true as per the syllabus, which explicitly mentions this fact. The reason states that according to Huygens’ principle, secondary wavelets propagate in all directions with the speed of the original light. This is also true as it is part of Huygens’ principle. However, the reason does not explain why rays are normal to the wavefront; it only describes the behavior of secondary wavelets. Therefore, both the assertion and reason are true, but the reason is not the correct explanation of the assertion.

Key Concept: Cylindrical Wavefront

c) Cylindrical
[Solution Description]
A linear (long narrow slit) source emits waves that propagate perpendicularly to its length. As a result, the wavefront takes the shape of a cylinder with the slit acting as the axis. Therefore, the wavefront is cylindrical.

Key Concept: Refraction of Light – Huygens’ Explanation

b) $\frac{v_1}{v_2}$
[Solution Description]
The refractive index of the second medium with respect to the first medium ($_1\mu_2$) is defined as the ratio of the speed of light in the first medium ($v_1$) to the speed of light in the second medium ($v_2$). Mathematically, it is given by $_1\mu_2 = \frac{v_1}{v_2}$.

Key Concept: Optical Path

b) $3.5D$
[Solution Description]
The optical path length in a medium is given by $n \times D$. For the first medium, optical path length is $1.5D$. For the second medium, it is $2.0D$. Total optical path length is the sum:
$1.5D + 2.0D = 3.5D$

Key Concept: Corpuscular Theory

d) Sir Isaac Newton
[Solution Description]
The corpuscular theory of light was proposed by Sir Isaac Newton in 1676.

Key Concept: Wavefronts, Optical Path

b) $\lambda$
[Solution Description]
Calculate optical paths separately:
Optical path in air = $n_1 \times D_1 = 1 \times 3\lambda = 3\lambda$.
Wavelength in medium = $\lambda/n_2 = \lambda/\sqrt{3}$.
Actual distance traveled in medium = $2\lambda = 2 \times (\lambda/\sqrt{3}) \times \sqrt{3}$, but this needs correction.
Optical path in medium = $n_2 \times D_2 = \sqrt{3} \times 2(\lambda/\sqrt{3}) = 2\lambda$.
Optical path difference = $3\lambda - 2\lambda = \lambda$.

Key Concept: Refraction and Snell's Law

a) The refractive indices of the two media
[Solution Description]
Huygens’ principle explains refraction using secondary wavelets. The ratio $\frac{\sin i}{\sin r}$ is equal to the ratio of the velocities of light in the two media ($\frac{v_1}{v_2}$), which is Snell’s law of refraction.

Key Concept: Wavefronts and Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when a plane wavefront is incident on a plane surface, the reflected wavefront remains planar. This is true because the angle of incidence equals the angle of reflection, preserving the wavefront's planar nature. The reason correctly explains the assertion by invoking Huygens’ principle: each point on the incident wavefront emits secondary wavelets, and their forward envelope forms the new wavefront. Since both statements are true and the reason logically justifies the assertion, the correct choice is option (a).

Key Concept: Speed of Light in Medium

b) $\frac{c}{2}$
[Solution Description]
The speed of light in a medium ($v$) is related to the speed of light in vacuum ($c$) and the refractive index ($n$) by the formula:
$v = \frac{c}{n} = \frac{c}{2}$
Thus, the speed of light in the medium is $\frac{c}{2}$.

Key Concept: Frequency remains constant

c) Frequency
[Solution Description]
When light travels from one medium to another, its frequency ($v$) remains constant. Only the speed and wavelength change as they depend on the refractive index of the medium.

Key Concept: Cylindrical Wavefront, Amplitude

b) $\frac{A_0}{\sqrt{2}}$
[Solution Description]
For a cylindrical wavefront, amplitude ($A$) varies as $A \propto \frac{1}{\sqrt{r}}$. If $A_0$ is the amplitude at distance $d$, then at distance $2d$, the amplitude $A$ is given by:
$A = A_0 \sqrt{\frac{d}{2d}} = A_0 \sqrt{\frac{1}{2}} = \frac{A_0}{\sqrt{2}}$

Key Concept: Huygens' Wave Theory

b) Luminiferous ether
[Solution Description]
Huygens proposed the existence of a hypothetical medium called 'luminiferous ether' as the medium for light propagation. This medium was believed to be weightless and capable of penetrating all matter while allowing light to travel at high speeds.

Key Concept: Huygens' Principle and Secondary Sources

b) All points on the wavefront act as secondary sources emitting spherical wavelets at the same speed.
[Solution Description]
According to Huygens’ principle, every point on a wavefront acts as a secondary source and emits wavelets in all directions with the speed of light in the medium. The new wavefront is the tangent to these secondary wavelets in the forward direction. This explains how wavefronts propagate through space.

Key Concept: Refraction using Huygens' Principle

c) The envelope of wavelets in medium 2 bends towards the normal if $v_2 < v_1$
[Solution Description]
Points on AB act as sources of secondary wavelets. In medium 2, the wavelet from A grows to radius $AB' = v_2 t$, while the wavelet from A' starts later. The common tangent A'B' represents the refracted wavefront. The ratio $\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$ yields Snell's law. Here, $i$ and $r$ are angles of incidence and refraction respectively.

Key Concept: Optical Path

c) Product of refractive index and physical distance
[Solution Description]
The optical path length is the product of the refractive index ($n$) of the medium and the physical distance ($D$) traveled by light in the medium, represented as $nD$. It accounts for the medium's effect on light propagation.

Key Concept: Reflection and Huygens' Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when a plane wavefront is incident on a plane reflecting surface, the reflected wavefront remains plane. This is correct because reflection from a plane surface preserves the planarity of the wavefront. The reason explains this using Huygens' principle, which describes how secondary wavelets generated by points on the incident wavefront result in a plane reflected wavefront. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets, which spread out in all directions with the same speed as the original wave. The new wavefront is formed by the tangent (envelope) to all these secondary wavelets in the forward direction. The reason correctly explains why the assertion is true. Both the assertion and reason are true, and the reason is the correct explanation of the assertion.

Key Concept: Optical Path, Refractive Index

d) 28 cm
[Solution Description]
The optical path length ($\text{OP}$) is given by:
$\text{OP} = nD$
where $n$ is the refractive index and $D$ is the geometric distance. Substituting the given values:
$\text{OP} = 1.4 \times 20 \, \text{cm} = 28 \, \text{cm}$

Key Concept: Reflection, Refraction, Wavefront Shape

d) Spherical converging wavefront
[Solution Description]
When a plane wavefront passes through a convex lens, it converges to a point at the focus. Therefore, the refracted wavefront will be a spherical converging wavefront.

Key Concept: Key Points of Huygens’ Theory

d) It contradicts experimental observations of reflection
[Solution Description]
Huygens’ principle suggests that secondary wavelets propagate both forward and backward. However, the backward wavefront contradicts experimental observations since no such backward-traveling light is observed during reflection. Hence, it is ignored.

Key Concept: Huygens’ Principle, Laws of Reflection

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Using Huygens’ principle, when the wavefront $AB$ strikes the surface $SS'$ at point $A$, each point on $AB$ acts as a source of secondary wavelets. The wavelet from $A$ has radius $AB' = BA' = vt$ after time $t$. Drawing a tangent $A'B'$ gives the reflected wavefront. From $\triangle ABA'$, $\sin i = \frac{BA'}{AA'}$, and from $\triangle AB'A'$, $\sin r = \frac{AB'}{AA'}$. Since $BA' = AB'$, we have $\sin i = \sin r$, proving $i = r$. The reason correctly explains the assertion as it links the equality of distances ($BA' = AB'$) to the equality of angles ($i = r$).

Key Concept: Wave theory basics

b) Waves
[Solution Description]
Huygens proposed that light propagates as waves, not particles. These waves require a medium called 'luminiferous ether' for their propagation, and they travel at the speed of light.

Key Concept: Snell's Law

b) $\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$
[Solution Description]
From the syllabus, Snell's law states that $\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = n$, where $i$ is the angle of incidence, $r$ is the angle of refraction, and $n$ is the refractive index of the second medium relative to the first.

Key Concept: Energy Transmission

c) It decreases inversely with the distance.
[Solution Description]
For a cylindrical wavefront, the energy spreads over the surface area of a cylinder, which is proportional to the radial distance ($r$). Therefore, the intensity ($I$) decreases inversely with the distance, i.e., $I \propto \frac{1}{r}$. The correct option must reflect this relationship.

Key Concept: Wavelength change, Refractive index relationship

a) $\frac{c}{\lambda}$
[Solution Description]
Frequency remains constant during refraction. The speed and wavelength change, but frequency ($v$) stays the same in both media. Thus, the frequency in the medium equals the frequency in air, which is given by $v = \frac{c}{\lambda}$ where $c$ is the speed of light in air.

Key Concept: Huygens' Principle - Reflection

b) Secondary wavelets are generated from each point on the wavefront
[Solution Description]
According to Huygens’ principle, each point on the incident wavefront acts as a source of secondary wavelets. When a plane wavefront strikes a reflecting surface, these secondary wavelets generate the reflected wavefront, allowing the derivation of the laws of reflection.

Key Concept: Huygens’ Explanation of Refraction

c) $\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$
[Solution Description]
According to Huygens’ explanation, when a plane wavefront passes from one medium to another, the ratio of the sines of the angles of incidence ($i$) and refraction ($r$) is equal to the ratio of the velocities of light in the two media. Mathematically, this is given by:
$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$
Since the refractive index $n$ is defined as $n = \frac{c}{v}$, where $c$ is the speed of light in vacuum, we can also write:
$\frac{v_1}{v_2} = \frac{n_2}{n_1}$
Thus, the correct relationship derived from Huygens' principle is $\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$.

Key Concept: Wavelength in denser medium, Huygens' Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Huygens' principle, the frequency ($v$) of light does not change when it moves from one medium to another. However, the speed ($v$) and wavelength ($\lambda$) decrease in a denser medium because the refractive index ($n$) is greater than 1. The formula for wavelength in the medium ($\lambda' = \frac{\lambda}{n}$) shows that wavelength decreases as $n$ increases. Therefore, Assertion (A) is true, as the wavelength decreases due to the decrease in speed. Reason (R) is also true, as frequency remains constant. Moreover, Reason (R) correctly explains Assertion (A) since the unchanging frequency ensures the relationship between wavelength and speed holds true.

Key Concept: Wavelength Comparison

b) It becomes half
[Solution Description]
Using the formula:
$\lambda' = \frac{\lambda}{n}$
Here, $\lambda = 500 \, \text{nm}$ and $n = 2$.
Substituting the values:
$\lambda' = \frac{500}{2} = 250 \, \text{nm}$
The wavelength in the medium ($250 \, \text{nm}$) is half of its wavelength in vacuum ($500 \, \text{nm}$).

Key Concept: Wavefront, Huygens’ Principle

d) Spherical, converging
[Solution Description]
When a plane wavefront is incident on a concave mirror, each point on the wavefront acts as a source of secondary wavelets. The reflected wavefront is formed by the envelope of these secondary wavelets. Due to the curvature of the mirror, the reflected wavefront becomes spherical, converging towards the focus of the mirror.

Key Concept: Huygens' Principle and Reflection

c) The reflected wavefront is formed by the envelope of secondary spherical wavelets
[Solution Description]
Huygens' principle states that every point on a wavefront acts as a source of secondary spherical wavelets. For reflection:
1. The incident wavefront AB strikes the reflecting surface at point A first.
2. Secondary wavelet from A spreads out with radius $v \Delta t$.
3. Reflected wavefront is tangent to these secondary wavelets.
4. Geometry shows angle of incidence $i$ equals angle of reflection $r$.

Key Concept: Snell's Law, Optical Path

c) $3.0 \, \text{cm}$
[Solution Description]
The optical path length is calculated using the formula $\text{Optical Path} = n \times D$, where $D$ is the physical distance. Here, $n = 1.5$ and $D = 2 \, \text{cm}$. Thus:
$\text{Optical Path} = 1.5 \times 2 = 3 \, \text{cm}$.

Key Concept: Huygens' Principle, Optical Path

d) 17 cm
[Solution Description]
The optical path length (OPL) is given by $n_1 D_1 + n_2 D_2$, where $D_1$ and $D_2$ are the distances travelled in each medium. Substituting the given values:
$OPL = (1.5 \times 6) + (2.0 \times 4) = 9 + 8 = 17$ cm.

Key Concept: Failure to Explain Refraction

a) Speed increases
[Solution Description]
The corpuscular theory predicted that the speed of light should increase in a denser medium like water or glass compared to air. However, experiments showed the opposite—light slows down in denser media. This contradiction was a major limitation of the corpuscular theory.

Key Concept: Speed and wavelength relationship

b) $\frac{\lambda}{3}$
[Solution Description]
The speed of light in the medium is $v = \frac{c}{3}$. The refractive index $n$ can be found as:
$n = \frac{c}{v} = \frac{c}{\frac{c}{3}} = 3$
Using the relation for wavelength:
$\lambda' = \frac{\lambda}{n} = \frac{\lambda}{3}$
The wavelength in the medium is one-third of its wavelength in vacuum.

Key Concept: Wavefront Types, Huygens’ Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that a plane wavefront becomes spherical upon reflection from a concave mirror, which is true because concave mirrors are converging optical elements and convert parallel rays (plane wavefront) into converging rays (spherical wavefront).
The Reason correctly explains this behavior by invoking Huygens’ principle, where secondary wavelets from the plane wavefront reconstruct the wavefront as spherical due to the curvature of the mirror. Thus, both are true and the Reason justifies the Assertion.

Key Concept: Huygens’ Principle and Reflection

d) Every point on the wavefront emits secondary wavelets, and the reflected wavefront is the tangent to these wavelets
[Solution Description]
Huygens' principle states that every point on a wavefront acts as a source of secondary wavelets. When a plane wavefront strikes a reflecting surface, each point on the surface emits secondary wavelets. The new reflected wavefront is formed by drawing a tangent to these secondary wavelets.

Key Concept: Effect of Refraction on Frequency

c) No, it remains unchanged
[Solution Description]
The frequency of light ($v$) depends solely on the source and does not change when light passes from one medium to another. Only the speed and wavelength of light change based on the refractive index of the medium.

Key Concept: Direction of propagation in isotropic media

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 in an isotropic medium is independent of the direction of propagation. This is true because isotropic media have uniform properties in all directions, including the refractive index ($n$). Since the speed of light in a medium ($v$) depends on the refractive index as $v = \frac{c}{n}$, and $n$ is constant in all directions for an isotropic medium, the speed $v$ remains unchanged regardless of the direction of propagation.
The reason explains that in an isotropic medium, the refractive index is the same in all directions, which is correct and directly supports the assertion. Therefore, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Optical Path

b) $nD$
[Solution Description]
The optical path length ($\text{OPL}$) is defined as the product of the refractive index ($n$) and the distance ($D$) traveled by light in the medium: $\text{OPL} = nD$.

Key Concept: Direction of Propagation

c) It is normal to the wavefront.
[Solution Description]
According to Huygens' principle, the direction of propagation (ray) is always perpendicular (normal) to the wavefront in a homogeneous isotropic medium. This ensures that the disturbance travels forward uniformly. Therefore, the correct answer must state this relationship.

Key Concept: Speed of sound, Refractive index

c) $\frac{74}{17}$
[Solution Description]
For sound waves, $\lambda = \frac{v}{f}$, where $v$ is speed and $f$ is frequency. Given same $f$, the wavelength ratio equals speed ratio:
$\frac{\lambda_{\text{water}}}{\lambda_{\text{air}}} = \frac{v_{\text{water}}}{v_{\text{air}}} = \frac{1480}{340} = \frac{74}{17}$

Key Concept: Energy Transmission

d) $I \propto \frac{1}{r^2}$
[Solution Description]
As per the syllabus, the intensity ($I$) of a spherical wave decreases with the square of the distance ($r$) from the point source. Mathematically, $I \propto \frac{1}{r^2}$.

Key Concept: Corpuscular Theory and Wave Theory

d) Assertion is false, but Reason is true.
[Solution Description]
According to the syllabus, Newton's corpuscular theory suggested light is made of particles (corpuscles). While it explained reflection and refraction, it failed to explain interference, diffraction, and polarization. The assertion falsely claims that corpuscular theory explained interference/diffraction. The reason correctly states that corpuscles cannot explain wave phenomena like interference. Thus, Assertion is false, but Reason is true.

Key Concept: Refraction Problems: Calculating Wavelength/Speed in Different Media

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The Assertion states that when light moves from a rarer to a denser medium, its wavelength decreases due to the decrease in speed. This is correct because according to the relationship $n = \frac{c}{v}$ and $n = \frac{\lambda}{\lambda'}$, as the refractive index ($n$) increases, both speed ($v$) and wavelength ($\lambda'$) decrease.
The Reason claims that the refractive index is inversely proportional to the speed of light in the medium, which is true as per $v = \frac{c}{n}$.
However, while both statements are true, the Reason does not directly explain why the wavelength decreases (the Assertion). The decrease in wavelength is due to the change in the medium's optical density, not merely the inverse proportionality between refractive index and speed.
Therefore, both Assertion and Reason are true, but the Reason does not correctly explain the Assertion.

Key Concept: Wavefront Shape

b) Spherical
[Solution Description]
A point source emits waves uniformly in all directions, resulting in spherical wavefronts as each point on the wavefront moves outward equally in all directions.

Key Concept: Huygens’ Principle, Wavefronts

c) 68°
[Solution Description]
Using Snell's law:
$\mu_1 \sin i = \mu_2 \sin r$.
Given $i = 30^\circ$, $\mu_1 = 1.5$, $\mu_2 = 2$:
$1.5 \sin 30^\circ = 2 \sin r \implies \sin r = \frac{1.5 \times 0.5}{2} = 0.375.$
Thus, $r = \sin^{-1}(0.375) \approx 22^\circ$.
The angle between the refracted wavefront and the boundary is $90^\circ - r \approx 68^\circ$.

Key Concept: Refraction and Wavelength

c) It decreases by a factor of 1.33.
[Solution Description]
The wavelength in a medium $\lambda_w$ is related to the wavelength in air $\lambda$ by $$\lambda_w = \frac{\lambda}{n}$$. Since $n > 1$, the wavelength decreases in water.

Key Concept: Huygens' Principle, Reflection

b) $90^\circ - \theta_i$
[Solution Description]
By Huygens' construction, the angle of incidence equals the angle of reflection ($\theta_i = \theta_r$). The wavefront is perpendicular to the propagation direction, so the angle between the reflected wavefront and the surface is the complement of $\theta_r$. Thus, it is $90^\circ - \theta_i$.

Key Concept: Properties of Ether

b) Low density and high elasticity
[Solution Description]
To account for the high speed of light ($3 \times 10^8 \, \text{ms}^{-1}$), Huygens assumed that the density of the luminiferous ether was very small and its elasticity was very large, as wave speed is given by $\text{wave speed} = \sqrt{\text{elasticity} / \text{density}}$.

Key Concept: Energy Transmission - Plane Wave Intensity

b) 50 W/m$^2$
[Solution Description]
For a plane wave, the intensity remains constant if there is no energy loss in the medium. Therefore, the intensity after traveling any distance will be the same as the initial intensity.

Key Concept: Inability to Explain Interference and Diffraction, No Explanation for Polarization

c) It contradicts the idea of light as particles; resolved by wave theory
[Solution Description]
The corpuscular theory couldn't explain interference patterns formed by light passing through double slits because such patterns are characteristic of wave superposition. Huygens' wave theory successfully explains this by treating light as a wave that interferes constructively and destructively.

Key Concept: Spherical Wavefront Propagation, Huygens’ Principle

b) $\frac{A_0}{2}$
[Solution Description]
The amplitude ($A$) of a spherical wavefront decreases with distance ($r$) as $A \propto \frac{1}{r}$. Given $A = A_0$ at $r = r_0$, we can write:
$A = \frac{k}{r}$
where $k$ is a constant. Substituting $A_0$ at $r_0$:
$A_0 = \frac{k}{r_0} \implies k = A_0 r_0$
Now, amplitude at $2r_0$:
$A = \frac{A_0 r_0}{2r_0} = \frac{A_0}{2}$

Key Concept: Wavefront Geometry

c) $BA' = AB'$
[Solution Description]
From the given explanation, both $BA'$ and $AB'$ are equal to $vt$, where $v$ is the velocity of light and $t$ is the time taken for the edge of the wavefront to reach the surface. Therefore, $BA' = AB'$.

Key Concept: Optical Path

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that the optical path is $nD$, which is correct as per the definition of optical path. The reason explains why this is the case: the optical path accounts for the time taken by light to travel distance $D$ in the medium, equivalent to the distance it would cover in vacuum in the same time. This matches the definition given in the syllabus. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

Key Concept: Snell’s Law

c) $19.47^\circ$
[Solution Description]
Using Snell’s law:
$\frac{\sin i}{\sin r} = \frac{n_2}{n_1}$
Given:
$n_1 = 1$,
$n_2 = 1.5$,
$i = 30^\circ$
Rearrange Snell’s law to find $r$:
$\sin r = \frac{n_1 \sin i}{n_2} = \frac{1 \cdot \sin 30^\circ}{1.5} = \frac{0.5}{1.5} = \frac{1}{3}$
Therefore:
$r = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ$

Key Concept: Huygens' Principle

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The Assertion states that each point on the wavefront $AB$ acts as a secondary source of wavelets, which is a direct application of Huygens' principle. The Reason explains Huygens' principle itself, stating that every point on a wavefront emits secondary wavelets. Since the Assertion directly follows from the Reason, and the Reason correctly explains why the Assertion is true, both are true and the Reason is the correct explanation of the Assertion.

Key Concept: Limitations of Corpuscular Theory

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The corpuscular theory, proposed by Newton, assumes light consists of tiny particles called corpuscles. However, interference is a phenomenon observed when two or more waves superpose to form a resultant wave. Since the corpuscular theory does not treat light as a wave, it cannot explain interference. Therefore, the Assertion is true. The Reason correctly identifies that interference is a wave property and the corpuscular theory treats light as particles, which is why it fails to explain interference. Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Refractive Index Definition

c) The ratio of the speed of light in medium 1 to that in medium 2: $\mu = \frac{v_1}{v_2}$
[Solution Description]
The refractive index ($\mu$) of medium 2 with respect to medium 1 is given by the ratio of the speed of light in medium 1 ($v_1$) to the speed of light in medium 2 ($v_2$):
$\mu = \frac{v_1}{v_2}$
This indicates how much slower light travels in medium 2 compared to medium 1.

Key Concept: Wavefront Propagation

c) The common tangent to the secondary wavelets in the forward direction
[Solution Description]
The new wavefront is determined by drawing the common tangent (envelope) to all the secondary wavelets emitted from points on the original wavefront after time $t$. The radius of each wavelet is given by $vt$, where $v$ is the velocity of light in the medium.

Key Concept: Plane wave intensity

c) It remains constant
[Solution Description]
A plane wave has a constant intensity as it propagates through a vacuum because its wavefront remains parallel and does not spread out. Thus, the energy per unit area per unit time does not change with distance.

Key Concept: Wavefront types, Phase difference

a) 0
[Solution Description]
For spherical wavefront, all points on same wavefront have same phase
Path difference between A and C = SC - SA = 2SA - SA = SA = 1 m
Phase difference = $(2\pi/\lambda)\times$ path difference
$\Delta\phi = (2\pi/0.5)\times 1 = 4\pi$ radians
Since $4\pi$ is equivalent to $2\times 2\pi$, there's effectively no phase difference

Key Concept: Ray Direction, Homogeneous Medium

c) Rays are normal to the wavefront
[Solution Description]
In a homogeneous isotropic medium, light rays are always normal (perpendicular) to the wavefront. This is because the medium's properties are uniform in all directions, causing the wavefront to propagate uniformly, and the rays (direction of energy propagation) align perpendicularly to the wavefront.
Step-by-step reasoning:
1. In a homogeneous isotropic medium, the speed of light is the same in all directions.
2. The wavefront propagates uniformly outward.
3. Rays represent the direction of energy flow, which is perpendicular to the wavefront in such a medium.
4. Thus, rays are always normal to the wavefront.

Key Concept: Huygens’ Wave Theory, Properties of ether

c) Both very small density and very large elasticity
[Solution Description]
According to Huygens' wave theory, the speed of a wave in a medium is given by $v = \sqrt{\text{elasticity} / \text{density}}$. For light to travel at a very high speed in the ether, the medium must have very small density and very large elasticity. Therefore, the correct answer is the combination of these two properties.
Step-by-step reasoning:
1. The formula for wave speed is $v = \sqrt{E/\rho}$, where $E$ is elasticity and $\rho$ is density.
2. High elasticity means the medium can quickly return to its original shape after distortion, allowing fast wave propagation.
3. Low density reduces resistance to wave motion, increasing speed.
4. Hence, both very large elasticity and very small density are required for high wave speed.

Key Concept: Construction of Forward Wavefront

c) $vt$
[Solution Description]
According to Huygens’ principle, each point on the wavefront acts as a source of secondary wavelets. After time $t$, these wavelets will have traveled a distance equal to $vt$. Therefore, the radius of each secondary wavelet is $vt$.

Key Concept: Huygens’ Principle, Wavefront

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
According to Huygens' principle, each point on a wavefront emits secondary wavelets that spread out spherically at the speed of light. The new wavefront is determined by the common tangent (envelope) of these secondary wavelets in the forward direction. Thus, both the Assertion and Reason are true, and the Reason correctly explains the Assertion as it describes the mechanism of wavefront propagation as per Huygens' principle.

Key Concept: Spherical Wavefront

c) Spherical wavefront
[Solution Description]
A point source emits waves that propagate equally in all directions, forming a spherical wavefront. All points on the surface of this sphere are at the same phase since the disturbance reaches them simultaneously.