CBSE Previous Year Questions: Wave Optics
CBSE Previous Year Questions 2025
Multiple Choice Type Questions
Reason (R): For interference, at least two waves are required. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Both Assertion (A) and Reason (R) are false.
Sol:
- The assertion is true because closing one slit results in a diffraction pattern from the single open slit.
- The reason is true because interference needs at least two waves.
- The reason correctly explains the assertion.
Hence, both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
Q2: Two coherent light waves, each having amplitude a, superpose to produce an interference pattern on a screen. The intensity of light as seen on the screen varies between: [1 Mark]
(A) 0 and 2a²
(B) 0 and 4a²
(C) a² and 2a²
(D) 2a² and 4a²
Ans: (B) 0 and 4a²
Sol: Let amplitudes of the two waves be .
$I \propto A^2$
Maximum intensity:
$I_{\max} = (a + a)^2 = (2a)^2 = 4a^2$
Minimum intensity:
$I_{\min} = (a - a)^2 = 0$
Therefore, intensity varies between $0$ and $4a^2$.
Very Short Answer Type Questions
Ans:
For single slit diffraction: \(a \sin \theta = n\lambda\)
First minimum: n=1
\( a \sin 30^\circ = 600\ \text{nm} \)
\( a \cdot \frac{1}{2} = 600 \Rightarrow a = 1200\ \text{nm} = 1.2\ \mu\text{m} \)
The correct Answer is: 1.2 μm
OR
Ans:
Phase difference: \(\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{8} = \frac{\pi}{4}\)
Resultant intensity:
\( I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \Delta \phi = 2I_0 + 2I_0 \cos(\pi/4) \)
\( I = 2I_0 (1 + 1/\sqrt{2}) = I_0 (2 + \sqrt{2}) \approx 3.414 I_0 \)
Q4: In a double-slit experiment, 6th dark fringe is observed at a certain point of the screen. A transparent sheet of thickness t and refractive index n is now introduced in the path of one of the two interfering waves to increase its phase by \(2\pi (n - 1)t/\lambda\). The pattern is shifted and 8th bright fringe is observed at the same point. Find the relation for thickness t in terms of n and λ. [2 Marks]
Ans:
$x_6 = \dfrac{11\lambda}{2}$
Final path difference for 8th bright fringe:
$x_8 = 8\lambda$
Extra path difference introduced:
$\Delta x = (n - 1)t$
Condition:
$x_6 + \Delta x = x_8$
$\dfrac{11\lambda}{2} + (n - 1)t = 8\lambda$
$(n - 1)t = 8\lambda - \dfrac{11\lambda}{2} = \dfrac{5\lambda}{2}$
$t = \dfrac{5\lambda}{2(n - 1)}$
The Correct Answer is: $t = \dfrac{5\lambda}{2(n - 1)}$
Q5: Find the angle of diffraction (in degrees) for first secondary maximum of the pattern due to diffraction at a single slit. The width of the slit and wavelength of light used are 0·55 mm and 550 nm, respectively. [2 Marks]
Ans:
For first secondary maximum:
\[ a \sin\theta = \frac{3\lambda}{2} \] \[ \sin\theta = \frac{3 \times 550 \times 10^{-9}}{2 \times 0.55 \times 10^{-3}} = \frac{1650 \times 10^{-9}}{1.1 \times 10^{-3}} = 1.5 \times 10^{-3} \] \[ \theta \approx \sin^{-1}(0.0015) \approx 0.086^\circ \]
Short Answer Type Questions
Q6: A double slit set-up was initially placed in a tank filled with water and the interference pattern was obtained using a laser light. When water is replaced by a transparent liquid of refractive index \(n > n_{\text{water}}\) what will be the effect on the following ? [3 Marks]
(a) Speed, frequency and wavelength of the light of laser beam.
(b) The fringe width, shape of interference fringes and shift in the position of central maximum.
(a) Speed, frequency, wavelength of laser light.
(b) Fringe width, shape of fringes, shift of central maximum.
Ans: (a)
- Speed of light will decrease
- Frequency remains unaffected
- Wavelength decreases
(b) Fringe width \(\beta = \frac{\lambda D}{d}\) decreases because \(\lambda\) decreases. Shape remains straight fringes. Central maximum does not shift because medium change is symmetric for both slits.
Q7: In Young's double slit experiment, the separation between the two slits is 1.0 mm and the screen is 1.0 m away from the slits. A beam of light consisting of two wavelengths 500 nm and 600 nm is used to obtain interference fringes. Calculate: [3 Marks]
(a) the distance between the first maxima for the two wavelengths.
(b) the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.
Ans:
(a) Position of nth bright fringe:
$x = \dfrac{n\lambda D}{d}$
For first order $(n = 1)$:
$x_1 = \dfrac{\lambda_1 D}{d}, \quad x_2 = \dfrac{\lambda_2 D}{d}$
Distance between them:
$\Delta x = \dfrac{D}{d} (\lambda_2 - \lambda_1)$
Given:
$D = 1,\text{m}, \quad d = 1,\text{mm} = 10^{-3},\text{m}$
$\Delta x = \dfrac{1}{10^{-3}} (600 - 500) \times 10^{-9}$
$= 10^3 \times 100 \times 10^{-9} = 10^{-4},\text{m} = 0.1,\text{mm}$
(b) Least distance where bright fringes coincide
Condition:
$n_1 \lambda_1 = n_2 \lambda_2$
$n_1 \times 500 = n_2 \times 600 \Rightarrow \dfrac{n_1}{n_2} = \dfrac{6}{5}$
Smallest integers: $n_1 = 6,; n_2 = 5$
Position:
$x = \dfrac{n_2 \lambda_2 D}{d} = \dfrac{5 \times 600 \times 10^{-9} \times 1}{10^{-3}} = 3 \times 10^{-3},\text{m} = 3,\text{mm}$
FAQs on CBSE Previous Year Questions: Wave Optics
| 1. What is wave optics? |  |
| 2. What is the principle of superposition in wave optics? | |
| 3. How does Young's double-slit experiment demonstrate wave properties of light? | |
| 4. What is diffraction and how does it occur? | |
| 5. What is polarisation in wave optics? | |
