Optics Question 1136
Question: In a Young’s double slit experiment, the slit separation is 1 mm and the screen is 1 m from the slit. For a monochromatic light of wavelength 500 nm, the distance of 3rd minima from the central maxima is
[Orissa JEE 2003]
Options:
A) 0.50 mm
B) 1.25 mm
C) 1.50 mm
D) 1.75 mm
Show Answer
Answer:
Correct Answer: B
Solution:
Distance of nth minima from central bright fringe $ x _{n}=\frac{(2n-1)\lambda D}{2d} $ .
For n=3 i.e. 3rd minima $ x _{3}=\frac{(2\times 3-1)\times 500\times {{10}^{-9}}\times 1}{2\times 1\times {{10}^{-3}}} $
$ =\frac{5\times 500\times {{10}^{-6}}}{2}=1.25\times {{10}^{-3}}m=1.25\ mm. $
