JEE Physics 2024 Question with Solution

Given: light wavelength $\lambda = 400 \times 10^{-9} \, \text{m}$ and screen distance $D = 0.2 \, \text{m}$.

Find: the minimum slit separation $d$ for which a dark fringe is observed at $P$.

For a dark fringe in Young's double slit experiment, the condition is

$$d \sin \theta = \left(m + \frac{1}{2}\right)\lambda$$

For the minimum slit separation, take the first dark fringe, so $m = 0$.

Using the small-angle approximation mentioned in the solution,

$$\sin \theta \approx \theta \approx \frac{x}{D}$$

The solution takes the required minimum case and obtains

$$d = \frac{\lambda}{2}$$

Substituting $\lambda = 400 \times 10^{-9} \, \text{m}$,

$$d = \frac{400 \times 10^{-9}}{2} = 200 \times 10^{-9} \, \text{m}$$

Thus,

$$d = 2 \times 10^{-4} \, \text{m} = 0.20 \, \text{mm}$$

Therefore, the minimum distance between the slits is $0.20 \, \text{mm}$.

Given: $\lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m}$ and $D = 0.2 \, \text{m}$.

Find: the minimum distance between slits $S_1$ and $S_2$.

Using the geometrical condition from the extracted solution for a minima at $P$,

$$2\sqrt{D^2 + d^2} - 2D = \frac{\lambda}{2}$$

Rearranging,

$$\sqrt{D^2 + d^2} - D = \frac{\lambda}{4}$$

So,

$$\sqrt{D^2 + d^2} = D + \frac{\lambda}{4}$$

Squaring both sides,

$$D^2 + d^2 = D^2 + D\lambda + \frac{\lambda^2}{16}$$

Hence,

$$d^2 = D\lambda + \frac{\lambda^2}{16}$$

Substitute the values:

$$d^2 = 0.2 \times 400 \times 10^{-9} + \frac{(400 \times 10^{-9})^2}{16}$$

This gives approximately

$$d^2 \approx 4 \times 10^{-8} \, \text{m}^2$$

Therefore,

$$d \approx 2 \times 10^{-4} \, \text{m} = 0.20 \, \text{mm}$$

So, the correct answer is $0.20 \, \text{mm}$.