In an experiment of a single slit diffraction, the width of the slit is 1.2 μm and the angular width of central maxima is observed to be equal to π/3. Find the wavelength of light.

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In an experiment of a single slit diffraction, the width of the slit is 1.2 μm and the angular width of central maxima is observed to be equal to π/3. Find the wavelength of light.

6 Å
60 Å
600 Å
6000 Å

Answer 1

CONCEPT:

Diffraction of light at single slit: In the case of diffraction at single slit, we get central bright band with alternate bright and dark bands.
- Width of the central maxima β₀ = 2λ D/d
- Angular width of central maxima b sin θ = λ
- All the secondary fringes are of same width but the central fringe having width double.
- Width of secondary fringes = λ D/d

F2 J.K 19.5.20 Pallavi D9

CALCULATION:

Given: Angular width of the central maximum = 2θ

2θ = π/3

⇒ θ = 30°

Width of the slit (b) = 1.2 μm = 1.2 × 10-6 m

θ = π/6 = 30° Width of the slit (b) = 1.2 μm = 1.2 × 10^-6 m

We know, b sin θ = n λ

For central maxima n = 1

⇒ b sin θ = λ

\(\lambda = 1.2 \times {10^{ - 6}} \times \sin 30^\circ = 1.2 \times {10^{ - 6}} \times \frac{1}{2} = 6 \times {10^{ - 7}}m = 6000 \times {10^{ - 10}}\;m \) = 6000 Å

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In an experiment of a single slit diffraction, the width of the slit is 1.2 μm and the angular width of central maxima is observed to be equal to π/3. Find the wavelength of light.

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