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NEET PHYSICSWave OpticsMedium

Question

In a diffraction pattern due to a single slit of width aa, the first minimum is observed at an angle 30∘30^\circ when light of wavelength 5000 A˚5000\text{ \AA} is incident on the slit. The first secondary maximum is observed at an angle of

A

sin⁡−1(23)\sin^{-1}\left(\frac{2}{3}\right)

B

sin⁡−1(12)\sin^{-1}\left(\frac{1}{2}\right)

C

sin⁡−1(34)\sin^{-1}\left(\frac{3}{4}\right)

D

sin⁡−1(14)\sin^{-1}\left(\frac{1}{4}\right)

Step-by-Step Solution

For a single slit diffraction pattern, the condition for the nthn^{\text{th}} minimum is given by asin⁡θ=nλa \sin\theta = n\lambda. For the first minimum (n=1n = 1), θ=30∘\theta = 30^\circ: asin⁡30∘=1⋅λa \sin 30^\circ = 1 \cdot \lambda a(12)=λ  ⟹  a=2λa \left(\frac{1}{2}\right) = \lambda \implies a = 2\lambda The condition for the nthn^{\text{th}} secondary maximum is asin⁡θ′=(n+12)λa \sin\theta' = \left(n + \frac{1}{2}\right)\lambda. For the first secondary maximum (n=1n = 1): asin⁡θ′=3λ2a \sin\theta' = \frac{3\lambda}{2} Substitute a=2λa = 2\lambda into the equation: (2λ)sin⁡θ′=3λ2(2\lambda) \sin\theta' = \frac{3\lambda}{2} sin⁡θ′=34\sin\theta' = \frac{3}{4} θ′=sin⁡−1(34)\theta' = \sin^{-1}\left(\frac{3}{4}\right)

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Wave Optics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWave Opticsdiffractionpatternsingleminimumobserved

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