JEE Physics 2025 Question with Solution

Given: Wavelength of light is $\lambda = 628 \, \text{nm}$. The angular separation between the second minimum on the left and the third minimum on the right is $30^\circ$.

Find: The slit width $a$.

For a single slit diffraction pattern, the condition for the $n^{\text{th}}$ minimum is

$$a \sin \theta_n = n\lambda$$

Using the small-angle approximation shown in the solution,

$$\theta_n \approx \frac{n\lambda}{a}$$

So for the second minimum,

$$\theta_2 \approx \frac{2\lambda}{a}$$

and for the third minimum,

$$\theta_3 \approx \frac{3\lambda}{a}$$

The total angular separation is the sum of these angular positions:

$$\Delta \theta = \theta_2 + \theta_3$$

Therefore,

$$\Delta \theta \approx \frac{2\lambda}{a} + \frac{3\lambda}{a} = \frac{5\lambda}{a}$$

So,

$$a \approx \frac{5\lambda}{\Delta \theta}$$

Convert $30^\circ$ into radians:

$$\Delta \theta = 30^\circ = \frac{\pi}{6}$$

Substitute the values:

$$a = \frac{5 \times 628 \times 10^{-9}}{\pi/6}$$ $$a = \frac{30 \times 628 \times 10^{-9}}{\pi}$$

Using the approximation stated in the solution, $628 \approx 200\pi$,

$$a \approx \frac{30 \times 200\pi \times 10^{-9}}{\pi}$$ $$a = 6000 \times 10^{-9} \, \text{m} = 6 \times 10^{-6} \, \text{m}$$

Hence,

$$a = 6 \, \mu\text{m}$$

Therefore, the width of the slit is $6 \, \mu\text{m}$, so the correct option is C.

Given: The second minimum lies to the left of the central maximum and the third minimum lies to the right. Their measured angular separation is $30^\circ$.

Find: The slit width $a$.

The solution also presents the minima positions as

$$\theta_1 = \sin^{-1}\left(\frac{2\lambda}{a}\right), \qquad \theta_2 = \sin^{-1}\left(\frac{3\lambda}{a}\right)$$

with

$$\theta_1 + \theta_2 = 30^\circ$$

So,

$$\sin^{-1}\left(\frac{2\lambda}{a}\right) + \sin^{-1}\left(\frac{3\lambda}{a}\right) = \frac{\pi}{6}$$

Solving this gives approximately

$$a = 6.07 \, \mu\text{m}$$

Thus the slit width is taken as

$$a \approx 6 \, \mu\text{m}$$

Therefore, the correct option is C.