Question

Monochromatic light of wavelength λ = 595 nm from a distant source passes through a slit 0.460 mm wide. The diffraction pattern is observed on a screen 4.00 m from the slit. In terms of the intensity I0 at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maximum?

a) 1.00mm

b) 3.00mm

c) 5.00 mm

Answer #1

Use these equations:

b = beta

t = theta

L = lambda

b = (2*pi*a*sin t)/L

I = I0*[[sin (b/2)]/(b/2)]^2

Since t is very very small,

sin t = tan t = y/x

b = 2*pi*a*y/(x*L)

b = 2*pi*4.6*10^-4*y/(4*595*10^-9) = 1214.39*y

Now Part A.

y = 1*10^-3

b = 1214.39*10^-3 = 1.214

b/2 = 0.607

I = I0*[[sin (0.607)]/0.607]^2 = 0.883*I0

Part B:

y = 3*10^-3

b/2 = 1214.39*3*10^-3/2 = 1.8215

I = I0*[[sin (1.8215)]/1.8215]^2 = 0.2828*I0

Part C

y = 5*10^-3

b/2 = 1214.39*5*10^-3/2 = 3.0359

I = I0*[[sin (3.0359)]/3.0359]^2 = 0.001207*I0

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