Diffraction

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A slit of width 'a' is illuminated by white light. The first minimum for red light ( $λ = 6500 \overset{0}{A}$ ) will fall at θ = 30⁰, then 'a' will be

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Solution

$d = a,\,\lambda = 6500{A^0} = 6500 \times {10^{ - 10}}m,\,n = 1\$
$\theta = {30^0},\sin \theta = \frac{{n\lambda }}{d},\,\sin {30^0} = \frac{{6500 \times {{10}^{ - 10}}}}{a}\$
$a = 13000 \times {10^{ - 10}}m = 1.3\,\,micron\$

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Diffraction

Remember concepts with our Masterclasses.

A slit of width 'a' is illuminated by white light. The first minimum for red light ( λ=6500 A0 ) will fall at θ = 300 , then 'a' will be

Ready to Test Your Skills?

free study material

A slit of width 'a' is illuminated by white light. The first minimum for red light ( $λ = 6500 \overset{0}{A}$ ) will fall at θ = 30⁰, then 'a' will be