1. Condition for minima in single-slit diffraction

For a single slit of width $a$, the diffraction minima occur at

$a\sin\theta_n = n\lambda, \qquad n=1,2,3,\dots$

where $n=1$ is the first minimum, $n=2$ is the second minimum, etc.

2. Use the given second minimum

The question says the second diffraction minimum is at $60^\circ$. So,

$a\sin 60^\circ = 2\lambda$

Since

$\sin 60^\circ = \frac{\sqrt{3}}{2},$

we get

$a\cdot \frac{\sqrt{3}}{2} = 2\lambda$

Hence,

$a = \frac{4\lambda}{\sqrt{3}}$

3. Now find the first minimum

For the first minimum,

$a\sin\theta_1 = \lambda$

Substitute $a=\dfrac{4\lambda}{\sqrt{3}}$:

$\frac{4\lambda}{\sqrt{3}}\sin\theta_1 = \lambda$

Cancel $\lambda$:

$\sin\theta_1 = \frac{\sqrt{3}}{4}$

4. Calculate $\theta_1$

$\theta_1 = \sin^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Now,

$\frac{\sqrt{3}}{4} \approx 0.433$

So,

$\theta_1 \approx \sin^{-1}(0.433) \approx 25.6^\circ$

Thus, $\theta_1$ is closest to

$\boxed{25^\circ}$

5. Check options

• A: $45^\circ$ ❌
• B: $30^\circ$ ❌
• C: $25^\circ$ ✅
• D: $20^\circ$ ❌

Therefore, the correct option is C.