CUET · PHYSICS · PYQ PAPER 2025
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 \( \frac{\pi}{3} \). If the first minimum is obtained at \( \theta_1 \), then \( \theta_1 \) is:
Given: \( \sin 25^\circ = \frac{\sqrt{3}}{4} \)
- A \( \frac{\pi}{3} \)
- B \( \frac{\pi}{4} \)
- C \( \frac{5\pi}{36} \)
- D \( \frac{\pi}{9} \)
Answer & Solution
Correct Answer
(C) \( \frac{5\pi}{36} \)
Step-by-step Solution
Detailed explanation
\(a \sin \theta = n \lambda\) \(a \sin (\frac{\pi}{3}) = 2 \lambda\) \(a \sin \theta_1 = \lambda\) \(\frac{\sin \theta_1}{\sin (\frac{\pi}{3})} = \frac{1}{2}\) \(\sin \theta_1 = \frac{1}{2} \sin (\frac{\pi}{3}) = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\)…
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