Class 12 Physics Chapter 10 Wave Optics

Key concept: Resultant Intensity

d) $$4I_0$$
[Solution Description] When constructive interference occurs, the intensity at that point is maximized and is given by $$4I_0$$, as all the wave crests or troughs align perfectly enhancing the overall intensity.

Key concept: Basic Concept

c) Thomas Young
[Solution Description] The scientist who contributed significantly to the wave theory of light, demonstrating that interference phenomena could be explained by wave properties, was Thomas Young.

Key concept: Spherical Wavefront

c) Spherical wavefront
[Solution Description] A spherical wavefront is generated by a point source because the waves emitted from such a source spread out uniformly in all directions, forming concentric spheres.

Key concept: Intensity Calculation

b) $$0$$
[Solution Description] The intensity at a point is calculated using the formula $$I = 4I_0 \cos^2 (\phi/2)$$. If the phase difference $$\phi$$ is $$\pi$$, then $$\cos(\pi/2) = 0$$. Therefore, $$I = 4I_0 \cdot 0 = 0$$.

Key concept: Total Internal Reflection

c) The angle of incidence is greater than the critical angle.

[Solution Description] Total internal reflection occurs when light moves from a denser medium to a rarer medium and the angle of incidence exceeds the critical angle. This means that the incident angle must be larger than the critical angle for total internal reflection to happen.

Key Concept: Geometrical Optics

c) Geometrical Optics

[Solution Description] Geometrical optics is a branch of optics that deals with the approximation where the wavelength of light is much smaller than the size of structures with which it interacts. In this approximation, light is treated as traveling in straight paths called rays. It does not consider wave effects such as diffraction and interference, which require considering light as having a finite wavelength. Therefore, by definition, geometrical optics neglects the wavelength.

Key Concept: Basic Concept

b) Principle of Superposition

[Solution Description] The principle responsible for explaining both diffraction and interference in light is the principle of superposition. Diffraction occurs when light bends around obstacles, and interference happens when two or more light waves overlap and combine. Both phenomena can be understood through the superposition principle, which states that the resultant displacement at any point is equal to the algebraic sum of the displacements due to each wave.

Key concept: Complex Pattern Analysis, Energy Conservation

c) The first minimum shifts outward and intensity decreases.

[Solution Description]

To determine how the position of the first minimum changes when the slit width is halved, we use the formula for diffraction minima: $$\theta = n\lambda/a$$, where $$a$$ is the slit width, $$n=1$$ for the first minimum, and $$\lambda$$ is the wavelength.

Initially, the angular position of the first minimum is calculated as:

$$\theta_1 = \frac{\lambda}{a}$$

When the slit width is reduced to half, i.e., $$a/2$$, the new angle for the first minimum becomes:

$$\theta_2 = \frac{\lambda}{a/2} = \frac{2\lambda}{a}$$

The angle $$\theta_2$$ is twice that of $$\theta_1$$. Thus, the position of the first minimum moves farther from the central maximum.

Regarding energy distribution, since the overall intensity is redistributed over a wider range due to increased angles for minima, the intensity at any given point decreases because light spreads out more.

Therefore, the change involves both an outward shift of the first minimum and a decrease in intensity.

Key concept: Light Intensity Change

c) The intensity is reduced by half.
[Solution Description] When unpolarised light passes through a single polaroid, its intensity is reduced by half. This is because the polaroid only allows the component of light vibrating in one plane to pass through, effectively reducing the intensity by 50%.

Key Concept: Critical Thinking, Real-World Problem Solving

d) 90°
[Solution Description]
Glare from horizontal surfaces like water is typically horizontally polarized. To reduce this glare maximally, the sunglasses should have vertical polarization axes, aligned perpendicular to the glare’s polarization direction. Typically, the ideal angle is around:
$$
\theta = 90^\circ – 0^\circ = 90^\circ.
$$
Thus, sunglasses should be vertically oriented or at approximately 90° relative to the horizontal plane.

Key concept: Huygens Principle Application

b) Plane
[Solution Description] According to Huygens principle, waves from a point source initially create spherical wavefronts. As these wavefronts travel further away from the source, they can be approximated as plane waves due to the curvature becoming negligible over large distances.

Key concept: Destructive Interference

a) 5 mm

[Solution Description] The position of a dark fringe is determined using the formula for destructive interference:

$$x_n = \left(n + \frac{1}{2}\right)\frac{D}{d} \lambda$$

where $$n = 0$$ for the first dark fringe. Substituting in the values:

\begin{align*}
x_{\frac{1}{2}} &= \left(0 + \frac{1}{2}\right) \frac{80 \times 10^{-2}}{0.04 \times 10^{-3}} \cdot 500 \times 10^{-9} \\
&= 0.5 \cdot 20000 \cdot 500 \times 10^{-9} \\
&= 5 \times 10^{-3} \, \text{m} \\
&= 5 \, \text{mm}
\end{align*}

Thus, the position of the first dark fringe is 5 mm.

Key concept: Practical Observation, Energy Conservation

c) Align the slit parallel to the bulb filament
[Solution Description] To clearly observe single slit diffraction patterns using a bulb filament, aligning the slit parallel to the filament aids in creating clearer diffraction bands. Adjusting the slit parallelism ensures minimal additional optical elements and maintains the conservation of energy by not altering the intrinsic properties of light being observed.

Key concept: Basic Concept

c) They have a constant phase difference and the same frequency.
[Solution Description] Coherent sources are defined as sources that have a constant phase difference and the same frequency. This means they maintain a stable relationship between their waves, allowing for predictable patterns of interference.

Key concept: Refraction Angle Calculation, Snell’s Law Application

b) $$62.73^\circ$$

[Solution Description] Using Snell’s Law $$n_1 \sin i = n_2 \sin r$$, where $$i = 45^\circ$$, we calculate $$\sin r$$:

$$\sin r = \frac{n_1 \sin i}{n_2} = \frac{1.5 \cdot \sin 45^\circ}{1.2}$$

Since $$\sin 45^\circ = \frac{\sqrt{2}}{2}$$,

$$\sin r = \frac{1.5 \cdot \frac{\sqrt{2}}{2}}{1.2} = \frac{1.5 \cdot 0.7071}{1.2} \approx 0.8839$$

Solving for $$r$$, $$r \approx \arcsin(0.8839) \approx 62.73^\circ$$.

Key concept: Phenomenon Explanation

d) It occurs only in transverse waves.

[Solution Description] Polarisation is a phenomenon that occurs only in transverse waves. In the case of light, which is an electromagnetic wave, polarisation refers to the orientation of the oscillations perpendicular to the direction of travel. Unlike longitudinal waves, where oscillations occur in the same direction as propagation, transverse waves can be polarised because their oscillations are perpendicular to the direction of motion.

Key concept: Basic Concept

a) $$q = n \lambda/a$$
[Solution Description]
Minima in a single slit diffraction pattern occur at angles where $$q = n \lambda/a$$, where $$n = \pm1, \pm2, \pm3, \ldots$$. This mathematical relation indicates that these are points of destructive interference.

Key Concept: Complex Refraction Scenarios, Real-world Application

c) Total internal reflection occurs within the glass

[Solution Description]

To solve this problem, we first apply Snell’s Law at the air-glass interface:

$$\sin \theta_1 = n_1 \sin \theta_2$$

where $$\theta_2$$ is the angle of refraction in the glass.

Next, when the light exits the glass into the medium with refractive index $$n_3$$, we again use Snell’s Law:

$$n_1 \sin \theta_2 = n_3 \sin \theta_3$$

Since $$n_3 < n_1$$, the angle $$\theta_3$$ will be greater than $$\theta_2$$, indicating the light bends away from the normal. Additionally, if $$\theta_3$$ becomes equal to or exceeds 90 degrees, total internal reflection can occur within the glass, causing all the light to reflect back rather than passing into the third medium. This happens if $$\theta_3 > \sin^{-1}(1/n_3)$$.

Therefore, depending on $$\theta_1$$, one possible outcome as the light exits into the third medium could be total internal reflection within the glass.

Key concept: Refraction Basics

c) Angle of incidence is less than angle of refraction

[Solution Description] According to Snell’s law, when light passes from a denser medium (higher refractive index) to a less dense medium (lower refractive index), the angle of refraction is greater than the angle of incidence.

Key concept: Critical Angle Calculation

c) 62.46°
[Solution Description]
The critical angle $$ i_c $$ can be calculated using the formula:
$$
\sin i_c = \frac{n_2}{n_1}
$$
where $$ n_2 = 1.33 $$ for water, and $$ n_1 = 1.5 $$ for glass.

Substituting the values:
$$
\sin i_c = \frac{1.33}{1.5} = 0.8867
$$

Find $$ i_c $$ by taking the inverse sine:
$$
i_c = \arcsin(0.8867) \approx 62.46^\circ
$$

Therefore, the critical angle is approximately $$ 62.46^\circ $$.

Key Concept: Experimental Design, Complex Problem Solving

a) $$I_0 \cos^2(15^\circ) \cdot \cos^6(30^\circ)$$
[Solution Description]
Start with the initial intensity $$I_0$$. Apply Malus’ Law successively for each polarizer:

After the first polarizer at 15 degrees:
$$
I_1 = I_0 \cos^2(15^\circ)
$$

After the second polarizer at 45 degrees (relative to previous):
$$
I_2 = I_1 \cos^2(45^\circ – 15^\circ) = I_1 \cos^2(30^\circ)
$$
So,
$$
I_2 = I_0 \cos^2(15^\circ) \cdot \cos^2(30^\circ)
$$

After the third polarizer at 75 degrees (relative to previous):
$$
I_3 = I_2 \cos^2(75^\circ – 45^\circ) = I_2 \cos^2(30^\circ)
$$
So,
$$
I_3 = I_0 \cos^2(15^\circ) \cdot \cos^4(30^\circ)
$$

Finally, after the fourth polarizer at 105 degrees (relative to previous):
$$
I_4 = I_3 \cos^2(105^\circ – 75^\circ) = I_3 \cos^2(30^\circ)
$$
Therefore,
$$
I_4 = I_0 \cos^2(15^\circ) \cdot \cos^6(30^\circ)
$$

Hence, the theoretical final intensity is $$I_0 \cos^2(15^\circ) \cdot \cos^6(30^\circ)$$.

Key Concept: Advanced Malus’ Law, Mathematical Derivation

b) $$\frac{I_0}{8}$$
[Solution Description] The first polaroid will polarize the light, reducing its intensity to half:

$$
I_1 = \frac{I_0}{2}
$$

According to Malus’ Law, the second polaroid further reduces this intensity based on the cosine squared of the angle between the pass axes:

$$
I_2 = I_1 \cos^2(60^\circ) = \frac{I_0}{2} \times \left(\frac{1}{2}\right)^2 = \frac{I_0}{8}
$$

Hence, the final intensity is $$\frac{I_0}{8}$$.

Key concept: Phase Difference

b) $$6\pi$$

[Solution Description]

The relationship between path difference ($$\Delta x$$) and phase difference ($$\Delta \phi$$) is given by:

$$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$$

Here, the path difference $$\Delta x = 3\lambda$$.

Substituting the values, we get:

$$\Delta \phi = \frac{2\pi}{\lambda} \times 3\lambda = 6\pi$$

Therefore, the phase difference is $$6\pi$$.

Key Concept: Integration of Concepts, Advanced Conceptual Application

b) Fringes become wider, and intensity decreases

[Solution Description] Halving the slit width $$d$$ affects the diffraction pattern according to the principle that the angular width of the central maximum $$\theta_c$$ is inversely proportional to $$d$$. The position of the central maximum $$x_0$$ on the screen for constructive interference will be given by $$x_0 = n \frac{D\lambda}{d}$$. When $$d$$ is halved, the width of the central maximum doubles because $$\theta_c = \frac{\lambda}{d/2}$$ becomes twice its original value. Therefore, the fringes become wider.

As for intensity, it is important to remember that intensity $$I$$ at the central maximum is related to the square of the amplitude, which is influenced by the slit width. When the slit width is reduced, the amplitude decreases, leading to a lower intensity at the central maximum.

Thus, with half the slit width:

– The position of the central maximum remains unchanged since it’s centered.

– The width of the central maximum increases, making the fringes wider.

– The intensity of the central maximum decreases due to the smaller aperture reducing the light passing through.

Therefore, the correct answer is: Fringes become wider, and intensity decreases.

Key Concept: Path Difference

a) 1.3 $$\mu m$$

[Solution Description]

The path difference $$\Delta x$$ for maxima in a double slit interference pattern is given by:

$$\Delta x = n \lambda$$

where $$n$$ is the order of the maximum and $$\lambda$$ is the wavelength of light used.

For the second order maximum ($$n = 2$$), the path difference is:

$$\Delta x = 2 \times 650 \, \text{nm}$$

To convert this into meters:

$$\Delta x = 2 \times 650 \times 10^{-9} \, \text{m}$$

$$\Delta x = 1300 \times 10^{-9} \, \text{m} = 1.3 \times 10^{-6} \, \text{m}$$

So, the path difference is $$1.3 \, \mu m$$.

Key Concept: Advanced Intensity Calculation, Young’s Experiment Analysis

c) $$I = 3I_0$$

[Solution Description] To find the resultant intensity at point $$P$$, we start with the formula for intensity in terms of phase difference $$\phi$$:

$$I = 4I_0 \cos^2 (\phi/2)$$

Given $$\phi = \frac{\pi}{3}$$, calculate $$\cos(\phi/2)$$:

$$\cos(\phi/2) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

Substituting this value into the intensity formula gives:

$$I = 4I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = 4I_0 \cdot \frac{3}{4} = 3I_0$$

Hence, the resultant intensity at point $$P$$ is $$3I_0$$.

Key Concept: Complex Snell’s Law, Multi-Medium Refraction

c) $$55.89^\circ$$

[Solution Description]

To find the final angle of refraction in medium 3, we need to apply Snell’s Law across each boundary.

**Step 1:** Calculate the angle of refraction in medium 2 using Snell’s Law:

$$n_1 \sin i = n_2 \sin r_2$$

where $$n_1 = 1.4$$, $$i = 50^\circ$$, and $$n_2 = 1.6$$.

Solve for $$\sin r_2$$:

$$\sin r_2 = \frac{n_1 \sin i}{n_2} = \frac{1.4 \times \sin(50^\circ)}{1.6}$$

The value of $$\sin(50^\circ)$$ is approximately 0.7660.

$$\sin r_2 = \frac{1.4 \times 0.7660}{1.6} \approx 0.6708$$

Thus, $$r_2 \approx \arcsin(0.6708) \approx 42.21^\circ$$.

**Step 2:** Use Snell’s Law again at the boundary between medium 2 and medium 3 to find $$r_3$$.

$$n_2 \sin r_2 = n_3 \sin r_3$$

Solve for $$\sin r_3$$:

$$\sin r_3 = \frac{n_2 \sin r_2}{n_3} = \frac{1.6 \times \sin(42.21^\circ)}{1.3}$$

Calculating $$\sin(42.21^\circ)$$ which is approximately 0.6711,

$$\sin r_3 = \frac{1.6 \times 0.6711}{1.3} \approx 0.8267$$

Therefore, $$r_3 \approx \arcsin(0.8267) \approx 55.89^\circ$$.

The angle of refraction when the light enters medium 3 is approximately $$55.89^\circ$$.

Key concept: Basic Definition

b) To ensure consistent and uniform wavefronts

[Solution Description] The primary purpose of using a coherent light source in holography is to ensure consistent and uniform wavefronts for accurate interference patterns. Coherence refers to the property of light where waves maintain a constant phase relationship. This consistency is crucial for creating clear and stable holograms. Without coherence, the overlapping light waves would not produce the sharp interference patterns needed for capturing three-dimensional images.

Key Concept: Crossed Polaroids

d) 0
[Solution Description] When two polaroid’s are crossed at 90°, no light passes through the second polaroid since the electric vectors are completely blocked. Hence, according to the syllabus, the transmitted intensity is zero.

Key concept: Basic Definition

a) A surface of constant phase

[Solution Description]

A wavefront is defined as a surface over which an optical wave has a constant phase. This means that all points on this surface are at the same phase of vibration; therefore, it represents the location where the wave propagates with uniform properties. Thus, option A correctly defines a wavefront.

Key concept: Effect of Filters on Diffraction Pattern

d) Green

[Solution Description] The visibility of fringes in the diffraction pattern depends on how well the colors are perceived by the human eye. Green light, with a wavelength around 550 nm, is close to the center of the visible spectrum and is more sensitive to human eyes, thus enhancing fringe visibility.

Key concept: Condition Recognition

d) $$x = (n+\frac{1}{2})\frac{D\lambda}{d}$$
[Solution Description] For destructive interference, leading to a dark fringe, the condition is:

$$
x = (n+\frac{1}{2})\frac{D\lambda}{d}
$$

This represents the path difference that causes cancellation.

Key concept: Coherent Sources

c) They should have a constant phase difference and the same frequency.
[Solution Description] Coherent sources are those which maintain a constant phase difference over time and have the same frequency, enabling stable interference patterns.

Key concept: Slit Width Effect

b) The central maximum becomes wider
[Solution Description] As the slit width ($$a$$) decreases, the width of the central maximum becomes wider. This is due to the inverse proportionality between slit width and the angular position of the minima, given by $$ \theta = n\lambda/a $$. Thus, for smaller $$a$$, $$\theta$$ increases, widening the central maximum.

Key concept: Complex Wavefront Analysis, Advanced Huygens’ Principle

c) 0

[Solution Description]

The focal length (f) of a concave mirror is given by the formula $$f = \frac{R}{2}$$. When a plane wave reflects off a concave mirror, it acts as if it’s converging at the focal point. Therefore, the distance between the focal point and the convergence point of the wavefront is zero because they meet exactly at the focal point.

Calculating the value:

\begin{equation}
f = \frac{R}{2}
\end{equation}

The converged wavefront meets at the focal point of the mirror itself, therefore:

\begin{equation}
\text{Distance to converge} = 0
\end{equation}

Hence, the correct answer is 0.

Key concept: Wave Equation Analysis

c) Linearly polarised transverse wave

[Solution Description] The wave equation given is of the form $$y(x,t) = A \cos(kx – \omega t)$$. This represents a plane polarised transverse wave because the displacement is perpendicular to the direction of wave propagation and oscillates in a single plane.

Key Concept: Real-World Application

c) They block horizontally polarized light, reducing glare.
[Solution Description]
Polarized sunglasses are designed to reduce glare by blocking horizontally polarized light. When light reflects off horizontal surfaces such as roads or water, it tends to become horizontally polarized. Polarized lenses have a vertical polarizing filter that only allows vertically oriented light waves to pass through, thereby significantly reducing glare and enhancing visual comfort and clarity.