Question

Deduce expression for angular width of central maximum

Answer 1

on the either side of the central maxima, there are first secondary minimas.

I hope it will help you

Answer 2

d \sin \theta = n \lambdadsinθ=nλ

n = 1, we have

d \sin \theta = \lambdadsinθ=λ

If angle is small, then \sin \theta = \thetasinθ=θ

d \theta = \lambdadθ=λ

Half angular width \theta = \frac {\lambda}{d}θ=

d

λ

Full angular width 2 \theta = 2 \frac {\lambda}{d}2 θ=2

d

λ

\omega' = \frac {2 \lambda}{d}ω

′

=

d

2λ

\frac {\lambda '}{\lambda} = \frac { \omega '} {\omega} \Rightarrow \lambda ' = \lambda ' \frac { \omega '} {omega}

λ

λ

′

=

ω

ω

′

⇒λ

′

=λ

′

omega

ω

′

\lambda ' = 6000 \times 0.7 = 4200 \mathring{A}λ

′

=6000×0.7=4200

A

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