Question

Consider in a Fraunhofer diffraction experiment at a single slit using the light of wavelength 4500 A, the first minimum is formed at an angle of 60°. The direction θ of the first secondary maximum is given by

(Consider \(\frac{\sqrt{3}}{2}\approx 0.9\))

1. \(sin^{-1}(0.45)\)
2. \(cos^{-1}(0.9)\)
3. \(sin^{-1}(0.9)\)
4. \(cos^{-1}(0.45)\)

Answer 1

Correct Answer - Option 1 : \(sin^{-1}(0.45)\)

CONCEPT:

• Diffraction: When a wave encounters an obstacle or opening in its way, the wave bends around the corners of an obstacle or through an aperture into the region of the geometrical shadow of the obstacle/aperture. This phenomenon is known as diffraction.

In diffraction for minima

b sinθ = nλ 

where b is the width of the slit, θ is the angle, λ is the wavelength, and n is the nth minima.

For maxima:

\(b sinθ = (n+ {1 \over 2}) λ\)

where b is the width of the slit, θ is the angle, λ is the wavelength, and n is the maxima number.

CALCULATION:

Given that λ = 4500 A

For first minima n = 1 and θ = 60°

b sinθ = nλ

b sin 60° = 1 × 4500 A

b = 5000 A

For maxima:

\(b sinθ = (n+ {1 \over 2}) λ\)

for 1st maxima n = 0

\(5000\times sinθ = (0+ {1 \over 2}) 4500\)

\(sinθ = 0.45\)

So the correct answer is option 3.