Question

In a single slit diffraction experiment, slit of width 'a' and incident light of wavelength \( 5600\,\text{\AA} \), the first minimum is observed at angle \( 30^\circ \). The first secondary maximum is observed at angle \( (\sin 30^\circ = 0.5) \)

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

Hint:

- First minimum gives slit width - Secondary maxima $\approx (m+\tfrac{1}{2})\lambda$

Answer 1

The Correct Option is D

Concept:
Single slit diffraction:
• Minima: \( a \sin\theta = m\lambda \), $m = 1,2,\dots$
• Secondary maxima (approx): \( a \sin\theta \approx \left(m+\frac{1}{2}\right)\lambda \)

Step 1: Use first minimum.
\[ a \sin 30^\circ = \lambda \;\Rightarrow\; a \cdot 0.5 = \lambda \;\Rightarrow\; a = 2\lambda \]

Step 2: First secondary maximum.
Occurs approximately at: \[ a \sin\theta \approx \frac{3}{2}\lambda \]

Step 3: Substitute $a = 2\lambda$.
\[ 2\lambda \sin\theta = \frac{3}{2}\lambda \;\Rightarrow\; \sin\theta = \frac{3}{4} \]

Step 4: Final angle.
\[ \theta = \sin^{-1}\left(\frac{3}{4}\right) \]