Wave Optics — NCERT Solutions

Given:

• Slit separation: $d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}$
• Distance to screen: $D = 1.4 \text{ m}$
• Distance between central fringe and 4th bright fringe: $y_4 = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}$

Concept: The position of the $n$-th bright fringe is given by:
$$y_n = \frac{n\lambda D}{d}$$

For the 4th bright fringe ($n = 4$):
$$y_4 = \frac{4\lambda D}{d}$$

Solving for $\lambda$:
$$\lambda = \frac{y_4 \cdot d}{4D}$$

$$\lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4}$$

$$\lambda = \frac{1.2 \times 0.28 \times 10^{-5}}{5.6}$$

$$\lambda = \frac{0.336 \times 10^{-5}}{5.6} = 0.06 \times 10^{-5} \text{ m}$$

$$\boxed{\lambda = 6 \times 10^{-7} \text{ m} = 600 \text{ nm}}$$

Given:

• At path difference $\Delta = \lambda$, intensity $I = K$
• Find intensity at path difference $\Delta' = \lambda/3$

Concept: The phase difference $\phi$ corresponding to a path difference $\Delta$ is:
$$\phi = \frac{2\pi}{\lambda} \cdot \Delta$$

The resultant intensity in Young's double slit experiment is:
$$I = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
where $I_0$ is the intensity due to each slit.

Step 1: Find $I_0$ using the condition at $\Delta = \lambda$:

Phase difference: $\phi_1 = \dfrac{2\pi}{\lambda} \times \lambda = 2\pi$

$$K = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0 \times (-1)^2 = 4I_0$$

So $I_0 = \dfrac{K}{4}$.

Step 2: Find intensity at $\Delta' = \lambda/3$:

Phase difference: $\phi_2 = \dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3} = \dfrac{2\pi}{3}$

$$I' = 4I_0 \cos^2\left(\frac{\phi_2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right)$$

$$I' = 4 \times \frac{K}{4} \times \left(\frac{1}{2}\right)^2 = K \times \frac{1}{4}$$

$$\boxed{I' = \frac{K}{4}}$$

Given:

• $\lambda_1 = 650 \text{ nm}$, $\lambda_2 = 520 \text{ nm}$
• Slit separation: $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ (standard assumed value for this problem)
• Distance to screen: $D = 1.2 \text{ m}$ (standard assumed value for this problem)

(Note: The problem does not specify $d$ and $D$; the answer is expressed in terms of $D/d$.)

(a) Distance of 3rd bright fringe for $\lambda_1 = 650$ nm:

Formula: $y_n = \dfrac{n\lambda D}{d}$

For $n = 3$, $\lambda_1 = 650 \text{ nm} = 650 \times 10^{-9} \text{ m}$:

$$y_3 = \frac{3 \times 650 \times 10^{-9} \times D}{d}$$

Using $D = 1.2$ m and $d = 2 \times 10^{-3}$ m:

$$y_3 = \frac{3 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$

$$y_3 = \frac{2340 \times 10^{-9}}{2 \times 10^{-3}} = 1170 \times 10^{-6} \text{ m}$$

$$\boxed{y_3 = 1.17 \times 10^{-3} \text{ m} \approx 1.17 \text{ mm}}$$

(b) Least distance where bright fringes of both wavelengths coincide:

Let the $n_1$-th bright fringe of $\lambda_1$ coincide with the $n_2$-th bright fringe of $\lambda_2$:

$$n_1 \lambda_1 = n_2 \lambda_2$$

$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1} = \frac{520}{650} = \frac{4}{5}$$

So the smallest integers satisfying this are $n_1 = 4$ and $n_2 = 5$.

Least distance (using $\lambda_1$, $n_1 = 4$):

$$y = \frac{n_1 \lambda_1 D}{d} = \frac{4 \times 650 \times 10^{-9} \times 1.2}{2 \times 10^{-3}}$$

$$y = \frac{3120 \times 10^{-9}}{2 \times 10^{-3}} = 1560 \times 10^{-6} \text{ m}$$

$$\boxed{y = 1.56 \times 10^{-3} \text{ m} = 1.56 \text{ mm}}$$

Verification: $y = \dfrac{5 \times 520 \times 10^{-9} \times 1.2}{2 \times 10^{-3}} = 1.56 \text{ mm}$ ✓