Question
Physics Question on Youngs double slit experiment
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5cm The distance between slits and screen is 1m and wavelength of used monochromatic light is 600mm The separation between the slits is:
A
12μm
B
60μm
C
36μm
D
48μm
Answer
48μm
Explanation
Solution
Given
D=1m
λ=600×10−9m
n=5
As ynth=dnλD
⇒d5×600×10−9×1=5×10−2
⇒d=5×10−25×600×10−9×1=60×10−6m
⇒d=60μm