### Neet Solved Paper 2013 Question 41

##### Question: In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $ {\lambda_1}=12000\overset{o}{\mathop{,A}}, $ and $ {\lambda_2}=10000\overset{o}{\mathop{,A}}, $ . At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

#### Options:

A) 8 mm

B) 6 mm

C) 4 mm

D) 3 mm

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

Given $ {\lambda_1}=12000\mathring {A} $ and $ {\lambda_2}=10000\mathring {A}, $

$ D=2cm $ and $ d=2,mm=2\times {10^{-3}}cm. $

We have $ \frac{{\lambda_1}}{{\lambda_2}}=\frac{h _1}{h _2} $ $ =\frac{12000}{10000}=\frac{6}{5} $

as $ x=\frac{u _1{\lambda_1}D}{d}=\frac{5\times 12000\times {10^{-10}}\times 2}{2\times {10^{-3}}} $

$ =5\times 1.2\times 10^{4}\times {10^{-10}}\times 10^{3}=6mm $