Solveeit Logo
Login

Question

Question: The light of wavelength 6328 Å is incident on a slit of width 0.2 mm perpendicularly, the angular wi...

The light of wavelength 6328 Å is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be.

A

0.36o0.36^{o}

B

0.18o0.18^{o}

C

0.72o0.72^{o}

D

0.09o0.09^{o}

Answer

0.36o0.36^{o}

Explanation

Solution

The angular half width of the central maxima is given by sin⁡θ=λa⇒θ=6328×10−100.2×10−3\sin\theta = \frac{\lambda}{a} \Rightarrow \theta = \frac{6328 \times 10^{- 10}}{0.2 \times 10^{- 3}}rad

=6328×10−10×800.2×10−3×π= \frac{6328 \times 10^{- 10} \times 80}{0.2 \times 10^{- 3} \times \pi}degree = 0.18o

Total width of central maxima =2θ=0.36o= 2\theta = 0.36^{o}