A monochromatic source of wavelength 60 nm was used in Young’s double slit experiment to produce interference pattern. I1 is the intensity of light at a point on the screen where the path difference is 150 nm. The intensity of light at a point where the path difference is 200 nm is given by

Question

A monochromatic source of wavelength 60 nm was used in Young’s double slit experiment to produce interference pattern. I1 is the intensity of light at a point on the screen where the path difference is 150 nm. The intensity of light at a point where the path difference is 200 nm is given by (A) (1/2) I1 (B) (3/2) I1 (C) (2/3) I1 (D) (3/4) I1

Tardigrade Admin · Accepted Answer

Given, λ=60 nm , Δ x1=150 nm and Δ x2=200 If the path difference is 150 nm then intensity I1=4 I0 cos 2 (φ/2) where, φ =(Δ x/λ) 2 π=(150/60) × 2 π=5 π ⇒ I1=4 I0 cos 2((5 π/2)) =4 I0 cos 2(2 π+(π/2)) ⇒ I1=0 Similarly, if path difference is 200 nm then intensity. I2=4 I0 cos 2 (φ/2) where, φ=(200/60) × 2 π=(20/3) π I2=4 I0 cos 2((20/3) π)=I0 So, I1=0 and I2=I0

$\frac{1}{2} I_{1}$

$\frac{3}{2} I_{1}$

$\frac{2}{3} I_{1}$

$\frac{3}{4} I_{1}$

$\frac{4}{3} I_{1}$

Q. A monochromatic source of wavelength 60nm was used in Young’s double slit experiment to produce interference pattern. I1​ is the intensity of light at a point on the screen where the path difference is 150nm. The intensity of light at a point where the path difference is 200nm is given by

21​I1​

23​I1​

32​I1​

43​I1​

34​I1​

$\frac{1}{2} I_{1}$

$\frac{3}{2} I_{1}$

$\frac{2}{3} I_{1}$

$\frac{3}{4} I_{1}$

$\frac{4}{3} I_{1}$