Light of wavelength 600 nm is incident normally on the slit of width 2 mm. What will be the angular width of central maxima at a distance of 1 m from the slit?

  1. 0.034°
  2. 0.34°
  3. 0.43°
  4. 0.043°

Answer (Detailed Solution Below)

Option 1 : 0.034°
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CUET General Awareness (Ancient Indian History - I)
10 Qs. 50 Marks 12 Mins

Detailed Solution

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CONCEPT:

  • When the monochromatic light ray falls on a single slit then it gets diffracted from the slit and form a bright and dark pattern on the screen.
  • The bright pattern is also called maxima and the dark pattern is called minima.
  • At maxima the intensity is maximum and at minima the intensity of light is minimum.

  • Here AB is the width of the central maxima/principal maxima and 2θ will be the angular width of principal maxima.

Where λ = wavelength of the light, n = an integer value, a = slit width, and D = distance of the screen from the slit.

Here θ is very small.

  • Sinθ ≈ θ

CALCULATION:

Given - λ = 600 nm = 600 × 10-9 m, D = 1 m and a = 2 mm = 2 × 10-3 m

The angular width of the central maxima (2θ) is

For principal maxima, n = 1

As rad = 180/π

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