Question

Below is a diffraction problem. I have the solutions to the problem but I don't know how to arrive at these solutions.

A) In an experiment two slits are separated by 0.22mm and illuminated by light of wavelength 640nm. How far must a screen be placed in order for the bright fringes to be separated by 5 mm? (1.72 m)

B) A soap film is illuminated by white light normal to its surface. The index of refraction of the film is 1.50. Light corresponding to a wavelength in air of 600 nm is intensified in the reflected beam. The smalled thickness of the film is.. (0.10 micrometers)

C) Unpolarized light of intensity Io passes through a pair of polarizing filters with their axes making an angle of 38 degrees. What fraction of Io passes through? (0.31)

Answer 1

all the formulas

pertain to double slit diffraction
condition for maxima is given by
y=m*lambda*D/d

where y=distance of m th order bright fringe from the center of screen
m=order of the bright fringe
lambda=wavelength of the light
D=distance of the screen
d=slit width

by putting m as any two consecutive integers, we can see that distance between any two bright fringes is given by

lambda*D/d

we are asked to find value of D.

by using all other known values, we get D=1.7188 m

b)in the relfected beam, for maximum reflection, the condition is

2*n*d*cos(theta)=(m-0.5)*lambda

where n=refractive index
d=thickness of film
theta=angle with vertical=0 degree (in this case)
lambda=wavelength

m=any integer, for lowest thickness, we need to take m=1

putting all avaialble values, we get d=1*10^(-7) m=0.1 micrometer

c)when it passes through first polarizing filter, as it is in the begin unpolairized, the light wave intensity will become 50%

so I1=I0/2

now when it passes thorugh second filter, according to Malus' law:

intensity will become I1*cos^2(38)

=(I0/2)*0.62=0.31*I0

so 31% of original intensity will pass through.