Discuss the special cases on first minimum in Fraunhofer diffraction.

1.	Discuss the special cases on first minimum in Fraunhofer diffraction.
Answer» Let us consider the condition for first minimum with (n = 1). a sin θ = λ The first minimum has an angular spread of, sin θ = \(\frac{λ}{a}\). Special cases to discuss on the condition. 1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1. 2. When a ≥ λ, the diffraction is possible. For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°. For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all. 3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac{λ}{a}\)= \(\frac{λ}{2λ}\) = \(\frac{1}{2}\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

Answer»

Let us consider the condition for first minimum with (n = 1). a sin θ = λ

The first minimum has an angular spread of, sin θ = \(\frac{λ}{a}\). Special cases to discuss on the condition.

1. When a < λ, the diffraction is not possible, because sin 0 can never be greater than 1.

2. When a ≥ λ, the diffraction is possible.

For a = λ, sin θ = 1 i.e, θ = 90°. That means the first minimum is at 90°. Hence, the central maximum spreads fully in to the geometrically shadowed region leading to bending of the diffracted light to 90°.
For a >> λ, sin θ << 1 i.e, the first minimum will fall within the width of the slit itself. The diffraction will not be noticed at all.

3. When a > λ and also comparable, say a = 2λ, sin θ = \(\frac{λ}{a}\)= \(\frac{λ}{2λ}\) = \(\frac{1}{2}\); then θ = 30°. These are practical cases where diffraction could be observed effectively.

Discuss the special cases on first minimum in Fraunhofer diffraction.

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