Question

In a single-slit Fraunhofer diffraction experiment, the first minimum is observed at an angle of \(30^{\circ }\) for light of wavelength \(\lambda\). If the same slit is now illuminated by light of the same wavelength, the angular position (\(\theta\)) of the first secondary maximum is
(\(\sin 30^{\circ }=\frac{1}{2}\)).

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

Hint:

Use the minima condition to find the slit width, then apply the approximate condition for the first secondary maximum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In single-slit diffraction the minima satisfy \(a\sin\theta = n\lambda\).
The first secondary maximum lies roughly midway between the first and second minima, so \(a\sin\theta = \dfrac{3\lambda}{2}\).

Step 2: Find the slit width
The first minimum is at \(30^{\circ}\), so \(a\sin 30^{\circ} = \lambda\).
This gives \(a\cdot\dfrac{1}{2} = \lambda\), so \(a = 2\lambda\).

Step 3: Calculation
\(\sin\theta = \dfrac{3\lambda}{2a} = \dfrac{3\lambda}{2\times 2\lambda} = \dfrac{3}{4}\).

Step 4: Final Result
Therefore \(\theta = \sin^{-1}\left(\dfrac{3}{4}\right)\).

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