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Monochromatic light of wavelength 441nmis incident on a narrow slit. On a screen 2.00maway, the distance between the second diffraction minimum and the central maximum is 1.50cm. (a) Calculate the angle of diffraction role="math" localid="1663145961694" θof the second minimum. (b) Find the width of the slit.

Short Answer

Expert verified

(a) The angle of diffraction for second order minima is 0.429°.

(b) The width of slit is 1.18×10-4m.

Step by step solution

01

Identification of given data

The wavelength of the monochromatic light is λ=441nm.

The distance of screen from narrow slit is D=2m.

The distance between central minima and second order minima is w=1.50cm.

The order of second minima is m=2.

02

Concept used

The diffraction is the phenomenon in which the light bends on the edges of the slit and pattern of light is formed on a screen.

03

Determination of angle of diffraction for second order minima

(a)

The angle of diffraction for second order minima is given as:

tanθ=wD

Substitute all the values in equation.

tanθ=1.50cm1m100cm2mθ=0.429°

Therefore, the angle of diffraction for second order minima is 0.429°.

04

Determination of width of slit

(b)

The width of slit is given as:

mλ=dsinθ

Substitute all the values in equation.

2441nm=dsin0.429°d=1.18×10-4m

Therefore, the width of slit is 1.18×10-4m.

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