### JEE Main On 8 April 2017 Question 4

##### Question: A single slit of width b is illuminated by a coherent monochromatic light of wavelength $ \lambda $ .If the second and fourth minima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e., distance between first minimum on either side of the central maximum) [JEE Online 08-04-2017]

#### Options:

A) 4.5 cm

B) 1.5 cm

C) 6.0 cm

D) 3.0 cm

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

- Min. $ f\sin \theta =n\lambda $ $ \sin \theta =\frac{n\lambda }{6} $ n = 2 $ \sin \theta =\frac{n\lambda }{6}=\tan {\theta_1}=\frac{x _1}{D} $ x = 4 $ \sin {\theta_2}=\frac{4\lambda }{6}=\frac{x _2}{D} $ $ x _2-x _1=\frac{4\lambda }{6}-\frac{2\lambda }{6}=\frac{2\lambda }{6}=3cm $ width of central $ \max =\frac{2\lambda }{6}=3cm $