Visible light of wavelength 600 nm falls normally on a single slit and diffraction pattern is obtained on a screen. It is found that the second diffraction minima is at \( (\pi/3) \). If the first minimum is obtained at \( \theta_{1} \), then \( \theta_{1} \) is
(Given: \( \sin 25^\circ = \sqrt{3}/4 \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( (5\pi/36) \)
In single slit diffraction the minima occur at \( a\sin\theta = n\lambda \), so \( \sin\theta \) is proportional to the order \( n \).
\( \displaystyle \sin\theta_{2} = \frac{2\lambda}{a} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \;\Rightarrow\; \frac{\lambda}{a} = \frac{\sqrt{3}}{4} \)
\( \displaystyle \sin\theta_{1} = \frac{\lambda}{a} = \frac{\sqrt{3}}{4} = \sin 25^\circ \)
\( \displaystyle \theta_{1} = 25^\circ = 25 \times \frac{\pi}{180} = \frac{5\pi}{36} \)