A diffraction pattern is formed on a screen 150 cm away from a 0.500-mm-wide slit. Monochromatic 546.1-nm light is used. Calculate the fractional intensity I/Imax at a point on the screen 4.10 mm from the center of the principal maximum.

Solution :

The expression for the intensity of light is given by :

For a small angle, θ

sin θ = tan θ

Therefore the above equation becomes,

The given data is

λ = 546.1 nm

L = distance between the slit and the screen = 140 cm

  = 1.40 m

a = width of the slit

  =

Therefore,

Therefore the fractional intensity is  

all of the answers are correct.

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explanation: