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Question: Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 mm . Calculate the wavelength of the light.

Short Answer

Expert verified

Light of 506 nm is used.

Step by step solution

01

Given Data

  • Width of slit, a=0.750mm=0.750×10-3m
  • Distance between screen and slit, R = 2.00 m
  • Distance of first minima from central maxima, y=1.35mm=1.35×10-3m
02

Concept

Diffraction is a process by which a beam of light spreads when passing through a narrow passage or across the edge of an obstacle accompanied by interference between the waveforms.

03

Wavelength of light source

By using the equation:

sinθ=mλaλ=asinθm

Here, a is the slit width, λis the wavelength of light used and is the diffraction angle.

For first minima, m = 1

λ=asinθ

To obtain the value of θ


tanθ=yRθ=tan-1yRθ=tan-11.35×10-3m2.0mθ=0.0387°

Therefore, the wavelength of source is given by

λ=asinθ=0.750×sin0.0387°=5.06×10-7m=506nm

Therefore, the wavelength of the light source is 506 nm.

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