Question

In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle \(30^{\circ }\) when light of wavelength \(5000\) Å is incident on the slit. The first secondary maximum is observed at an angle of
\(\{\sin (30)^{\circ }=\cos (60)^{\circ }=\frac{1}{2}\}\)

• A. \(\sin ^{-1}\left(\frac{2}{3}\right)\)
• B. \(\sin ^{-1}\left(\frac{1}{2}\right)\)
• C. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
• D. \(\sin ^{-1}\left(\frac{1}{4}\right)\)

Hint:

Use the first minimum condition to find the slit width in terms of the wavelength, then apply the condition for the first secondary maximum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a single slit, minima occur at \(a\sin\theta=n\lambda\) and secondary maxima occur approximately at \(a\sin\theta=\left(n+\tfrac{1}{2}\right)\lambda\).

Step 2: Find the slit width
The first minimum is at \(30^{\circ}\), so \(a\sin 30^{\circ}=\lambda\). Since \(\sin 30^{\circ}=\tfrac{1}{2}\), we get \(a=2\lambda\).

Step 3: Calculation
For the first secondary maximum, \(a\sin\theta=\tfrac{3\lambda}{2}\).
\(\sin\theta=\dfrac{3\lambda}{2\times 2\lambda}=\dfrac{3}{4}\)
So \(\theta=\sin^{-1}\left(\dfrac{3}{4}\right)\).

Final Answer:
The first secondary maximum is at the angle whose sine is 3/4, which is option (C). \[ \boxed{\text{(C) }\sin^{-1}\left(\dfrac{3}{4}\right)} \]