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} ) $
Show Answer
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} ) $
