In Young's double slit experiment, the wavelength of light used is ' '. The intensity at a point is I where path difference is ( 4 ). If I_0 denotes the maximum intensity, then the ratio ( II_0 ) is…

Question

In Young's double slit experiment, the wavelength of light used is ' '. The intensity at a point is I where path difference is ( 4 ). If I_0 denotes the maximum intensity, then the ratio ( II_0 ) is ( sin ⁡ 4= cos ⁡ 4= 1 2)

Accepted Answer

Correct answer: B. (b): Given, wavelength = , x= /4 I =I_0 cos^2 ⁡ ( 2 ) ; = 2 r= 2 , 4= 2 ∴I=I_0 cos^2 ⁡ ( 4 )=I_0 ( 1 2 )^2= I_02

$In Young's double slit experiment, the wavelength of light used is ' λ'. The intensity at a point is I where path difference is (\frac{λ}{4}) . If I_{0} denotes the maximum intensity, then the ratio (\frac{I}{I _{0}}) is (sin \frac{π}{4} = cos \frac{π}{4} = \frac{1}{2})$

Solution

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