JEE MAIN - Physics (2022 - 28th June Evening Shift - No. 16)
In Young's double slit experiment performed using a monochromatic light of wavelength $$\lambda$$, when a glass plate ($$\mu$$ = 1.5) of thickness x$$\lambda$$ is introduced in the path of the one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be :
3
2
1.5
0.5
Explanation
For the intensity to remain same the position must be of a maxima so path difference must be n$$\lambda$$ so
(1.5 $$-$$ 1) x$$\lambda$$ = n$$\lambda$$
x = 2n (n = 0, 1, 2 ....)
So, value of x will be
x = 0, 2, 4, 6 ....
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