Wave Optics

Here are 30 high-order thinking (HOTS) questions for CBSE Class 12 Physics Chapter "Wave Optics," with full solutions, including numericals, derivations, critical thinking, and conceptual explanations.

1. Question:

Derive an expression for the fringe width in Young's double-slit experiment and explain the effect of wavelength and slit separation on the fringe width.

Solution: The fringe width (β) in Young’s double-slit experiment is the distance between two consecutive bright or dark fringes. The formula for fringe width is given by:

$$\beta = \frac{\lambda D}{d}$$

Where:

• $\lambda$ is the wavelength of the light,
• $D$ is the distance between the screen and the slits,
• $d$ is the separation between the slits.

Effect of wavelength and slit separation:

• If the wavelength ($\lambda$ ) increases, the fringe width increases, i.e., the fringes become wider.
• If the slit separation ($d$ ) increases, the fringe width decreases, i.e., the fringes become closer.

2. Question:

A monochromatic light of wavelength 600 nm is used in an interference experiment. If the fringe width is 0.3 mm and the distance between the screen and slits is 2 m, calculate the slit separation.

Solution: Using the formula for fringe width:

$$\beta = \frac{\lambda D}{d}$$

Substituting the known values:

$$0.3 \times 10^{-3} = \frac{600 \times 10^{-9} \times 2}{d}$$

Solving for $d$ :

$$d = \frac{600 \times 10^{-9} \times 2}{0.3 \times 10^{-3}} = 4 \times 10^{-4} \, \text{m} = 0.4 \, \text{mm}$$

3. Question:

Explain the concept of coherence in wave optics. Why is it necessary for two light sources to be coherent for producing an interference pattern?

Solution: Coherence refers to the correlation between the phases of two or more waves. For two light waves to produce a stable interference pattern, they must be coherent. This means their phase difference should remain constant over time.

Coherence is necessary because if the phase difference changes randomly, the constructive and destructive interference will keep changing, and no stable interference pattern will form.

4. Question:

In a Young's double-slit experiment, the separation between the slits is 0.25 mm and the distance between the screen and the slits is 1.5 m. If the wavelength of the light used is 500 nm, calculate the fringe width.

Solution: Using the fringe width formula:

$$\beta = \frac{\lambda D}{d}$$

Substituting the given values:

$$\beta = \frac{500 \times 10^{-9} \times 1.5}{0.25 \times 10^{-3}} = 3 \, \text{mm}$$

5. Question:

In a Young's double-slit experiment, if the distance between the slits is reduced to half, how will the fringe width change? Explain.

Solution: From the formula for fringe width:

$$\beta = \frac{\lambda D}{d}$$

If the slit separation $d$ is reduced to half, the fringe width will double. This is because $\beta \propto \frac{1}{d}$ .

6. Question:

Explain the principle of superposition of waves with reference to interference. How does it lead to constructive and destructive interference?

Solution: The principle of superposition states that when two or more waves meet, the resultant displacement at any point is the algebraic sum of the displacements of the individual waves at that point.

• Constructive interference occurs when the waves are in phase, i.e., their displacements add up, resulting in a larger amplitude.
• Destructive interference occurs when the waves are out of phase, i.e., their displacements cancel each other, resulting in zero or reduced amplitude.

7. Question:

A thin film of air with refractive index 1.5 is placed between two glass plates. What is the condition for constructive interference in the reflected light?

Solution: For constructive interference in thin films, the condition is:

$$2 t = m \lambda_{\text{eff}}$$

Where $t$ is the thickness of the film and $\lambda_{\text{eff}} = \frac{\lambda}{n}$ is the effective wavelength of light in the medium, where $n$ is the refractive index of the film.

The condition becomes:

$$2 t = m \frac{\lambda}{n}$$

8. Question:

Derive the expression for the fringe shift when the medium between the slits and the screen is replaced by a medium of refractive index $n$ .

Solution: When the medium is replaced with one having refractive index $n$ , the wavelength in the medium becomes:

$$\lambda_{\text{new}} = \frac{\lambda}{n}$$

The new fringe width ($\beta_{\text{new}}$ ) is given by:

$$\beta_{\text{new}} = \frac{\lambda_{\text{new}} D}{d} = \frac{\lambda D}{n d}$$

So, the fringe width decreases by a factor of $n$ .

9. Question:

Explain why a sodium lamp produces a stable interference pattern despite being an incoherent source.

Solution: A sodium lamp emits light with a narrow spectral range (monochromatic light), which gives a quasi-coherent source. The light waves emitted by the sodium lamp have a fixed frequency, and the phase difference between them remains almost constant over a short time interval, which allows interference to occur.

10. Question:

In a Michelson interferometer, if the mirror is moved by a distance $d$ , how many fringes are shifted?

Solution: In a Michelson interferometer, if the mirror is moved by a distance $d$ , the path difference changes by $2d$ (because light travels to the mirror and back). For one fringe shift, the path difference must change by one wavelength ($\lambda$ ).

The number of fringes shifted is:

$$\text{Number of fringes} = \frac{2d}{\lambda}$$

11. Question:

In a diffraction grating, if the number of lines per meter is doubled, how does the angular separation of adjacent maxima change?

Solution: The angular separation of adjacent maxima in a diffraction grating is given by:

$$\sin \theta = \frac{m\lambda}{d}$$

Where $d$ is the grating spacing ($d = \frac{1}{N}$ , where $N$ is the number of lines per meter).

If the number of lines per meter is doubled, the spacing $d$ is halved, which results in the angular separation $\theta$ doubling.

12. Question:

A single slit diffraction pattern is observed. If the width of the slit is halved, how does the angular position of the first diffraction minimum change?

Solution: The angular position of the first diffraction minimum in a single-slit diffraction is given by:

$$\sin \theta = \frac{\lambda}{a}$$

Where $a$ is the slit width. If the slit width is halved, the angular position $\theta$ will increase, as $\sin \theta$ becomes larger. Therefore, the minimum moves further from the central maximum.

13. Question:

Explain the concept of diffraction grating and derive the condition for maxima in a diffraction pattern.

Solution: Diffraction grating consists of multiple slits. The condition for maxima in the diffraction pattern is given by:

$$d \sin \theta = m \lambda$$

Where $d$ is the distance between slits, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the incident light.

14. Question:

A monochromatic light of wavelength 500 nm passes through a single slit of width 0.1 mm. Calculate the angular position of the first diffraction minimum on the screen placed 2 meters away.

Solution: For single-slit diffraction, the angular position of the first minimum is given by:

$$\sin \theta = \frac{\lambda}{a}$$

Substituting the values:

$$\sin \theta = \frac{500 \times 10^{-9}}{0.1 \times 10^{-3}} = 5 \times 10^{-3}$$

Therefore:

$$\theta \approx \sin^{-1} (5 \times 10^{-3}) \approx 0.29^\circ$$

15. Question:

Derive the expression for the angular width of the central maximum in a single-slit diffraction pattern.

Solution: The angular width of the central maximum is the angular separation between the first minima on both sides of the central maximum. The position of the first minima is given by:

$$\sin \theta = \frac{\lambda}{a}$$

The angular width of the central maximum is twice this angle:

$$\Delta \theta = 2 \times \theta = 2 \sin^{-1} \left( \frac{\lambda}{a} \right)$$

16. Question:

In an interference experiment, the intensity at the center of the screen is 4 times the intensity at the first minimum. Find the ratio of the amplitudes of the two interfering waves.

Solution: The intensity in an interference pattern is given by:

$$I = I_0 (1 + \cos \delta)$$

At the first minimum, $\cos \delta = -1$ , so the intensity at the first minimum is:

$$I_{\text{min}} = I_0 (1 - 1) = 0$$

At the center (maximum), $\cos \delta = 1$ , so the intensity is:

$$I_{\text{max}} = I_0 (1 + 1) = 2 I_0$$

Given that the intensity at the center is 4 times the intensity at the first minimum, the intensity ratio is 4. Therefore, the amplitude ratio is:

$$\text{Amplitude ratio} = \sqrt{4} = 2$$

17. Question:

Calculate the angular separation of the second-order maxima in a diffraction grating with 3000 lines/cm and light of wavelength 600 nm.

Solution: The condition for maxima in a diffraction grating is:

$$d \sin \theta = m \lambda$$

For the second-order maximum ($m = 2$ ):

$$d = \frac{1}{3000 \times 10^2} \, \text{cm} = \frac{1}{3 \times 10^5} \, \text{m}$$

Substituting the values:

$$\sin \theta = \frac{2 \times 600 \times 10^{-9}}{ \frac{1}{3 \times 10^5}} = 0.36$$

Therefore:

$$\theta = \sin^{-1} (0.36) = 21^\circ$$

The angular separation between two second-order maxima is $2 \times 21^\circ = 42^\circ$ .

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