A) what is the distance ( in cm) on the screen from the center of this central m

Question

A) what is the distance ( in cm) on the screen from the center of this central maximum to the point where the intensity due to douple-slit interference 25% of I0?
B) what fraction of the maximum intensity on the screen is the intensity measured at a distance 2 mm from the central maximum? Problem 2 (20 pts) Light of wavelength 675 nm passes through two slits separated by 0.24 mm and produces an interference pattern on a screen 4.70 m away. The intensity at the central maximum is o a. What is the distance (in cm) on the screen from the center of this central maximum to the where the intensity due to double-slit interference 25% of 10? point b. What fraction of the maximum intensity on the screen is the intensity measured at a distance 2 mm from the central maximum?

Explanation / Answer

for any point at a distance of y from central maxima , if angle made with central axis is theta,

then tan(theta)=y/D

where D=distance of the screen from the slit

intensity of double slit interference pattern is given by

I=I0*(cos(pi*d*sin(theta)/lambda))^2

where d=slit width

theta=angle made with the central axis

lambda=wavelength

let at a distance of y , intensity is 1/4 th of I0

then 1/4=(cos(pi*d*sin(theta)/lambda))^2

==>cos(pi*d*sin(theta)/lambda)=1/2

==>pi*d*sin(theta)/lambda=pi/3

d*sin(theta)/lambda=1/3

==>sin(theta)=(1/3)*lambda/d

using the values as given in question,

lambda=675*10^(-9) and d=0.24*10^(-3) m

we get sin(theta)=9.375*10^(-4)

==>tan(theta)=9.375*10^(-4)

then distance from central maxima=4.7*9.375*10^(-4)=4.40625*10^(-3) m=0.04406 cm

part b:

at a distance of 2 mm, tan(theta)=0.002/4.7=4.25532*10^(-4)

==>sin(theta)=4.25532*10^(-4)

then intensity=I0*(cos(pi*d*sin(theta)/lambda))^2

=I0*(cos(pi*0.24*0.001*4.25532*10^(-4)/(675*10^(-9)))^2=0.79*I0

hence 79% of maximum intensity is observed at a distance of 2 mm

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