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Question: A parallel beam of light of wavelength 6000Å get diffracted by a single slit of width 0.3 mm. The an...

A parallel beam of light of wavelength 6000Å get diffracted by a single slit of width 0.3 mm. The angular position of the first minima of diffracted light is :

A

2×10-3 rad

B

3×10-3 rad

C

1.8×10-3 rad

D

6×10-3 rad

Answer

2×10-3 rad

Explanation

Solution

Here, λ=6000A˚=6000×10−10m=6×10−7m\lambda = 6000Å = 6000 \times 10^{- 10}m = 6 \times 10^{- 7}m

a=0.3mm=0.3×10−3m=3×10−4ma = 0.3mm = 0.3 \times 10^{- 3}m = 3 \times 10^{- 4}m

For first minima

asin⁡θ=λa\sin\theta = \lambda where a is the slit width

sin⁡θ=λa=6×10−7m3×10−4m=2×10−3\sin\theta = \frac{\lambda}{a} = \frac{6 \times 10^{- 7}m}{3 \times 10^{- 4}m} = 2 \times 10^{- 3}

As sin⁡θ\sin\theta is very small

∴θ≅sin⁡θ=2×10−3rad\therefore\theta \cong \sin\theta = 2 \times 10^{- 3}rad