NEET Solved Paper 2016 Question 6

Question: In a diffraction pattern due to a single slit of width ‘a’, the first minimum is observed at an angle $ 30^{o} $ when light of wavelength $ 5000\overset{o}{\mathop{A}}, $ is incident on the slit. The first secondary maximum is observed at an angle of :

Options:

A) $ {{\sin }^{-1}}( \frac{1}{4} ) $

B) $ {{\sin }^{-1}}( \frac{2}{3} ) $

C) $ {{\sin }^{-1}}( \frac{1}{2} ) $

D) $ {{\sin }^{-1}}( \frac{3}{4} ) $

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Answer:

Correct Answer: D

Solution:

For first minima, $ \sin 30^{o}=\frac{\lambda }{a}=\frac{1}{2} $

First secondary maxima will be at $ \sin \theta =\frac{3\lambda }{2a}=\frac{3}{2}( \frac{1}{2} )\Rightarrow \theta ={{\sin }^{-1}}( \frac{3}{4} ) $