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. $



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