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{\AA} $ and $ {\lambda_2}=10000{\AA}, $ $ 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 $
