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?

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.
• \(\sin \theta =\frac{n\lambda }{a}\)

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

\({\rm{Angular\;width\;}}\left( {2\theta } \right) = \frac{{2n\lambda }}{a}\)

For principal maxima, n = 1

\(\Rightarrow {\rm{Angular\;width\;}}\left( {2\theta } \right) = \frac{{2 \times 600 \times {{10}^{ - 9}}}}{{2 \times {{10}^{ - 3}}{\rm{\;}}}} = 6 \times {10^{ - 4}}rad\)

As rad = 180/π