Question

First diffraction minima due to a single slit of width \(10^{-4}\) cm is at \(\theta = 30^\circ\). Then wavelength of the light used is

• A. 4000 \AA
• B. 5000 \AA
• C. 6000 \AA
• D. 6250 \AA

Hint:

Single slit minima: \(a\sin\theta = n\lambda\) (n = 1, 2, ...). First minimum: \(n=1\). Remember \(1\ \AA = 10^{-10}\) m.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For first minima in single slit: \(a\sin\theta = \lambda\).
Step 2: Detailed Explanation:
\(a = 10^{-4}\) cm \(= 10^{-6}\) m
\[ \lambda = a\sin\theta = 10^{-6} \times \sin 30^\circ = 10^{-6} \times 0.5 = 5 \times 10^{-7} \text{ m} = 5000 \text{ \AA} \]
Step 3: Final Answer:
Wavelength \(= \mathbf{5000}\) \AA.