Question
Download Solution PDFLight 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCONCEPT:
- 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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