A parallel beam of monochromatic light of wavelength 5000 A is incident on a single narrow slit of width 0.001 mm. The light is focussed by a convex lens on a screen placed on focal plane. The first minimum will be formed formed for the angle of diffraction equal to

20^{\circ}

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15^{\circ}

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30^{\circ}

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50^{\circ}

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Solution

The correct option is A $20^{\circ}$
$d sin θ = m λ$
$θ = {sin}^{- 1} (\frac{m λ}{d})$
$= {sin}^{- 1} (\frac{(1) (5000 \times 10^{- 10})}{0.001 \times 10^{- 3}})$
$θ = 30^{o}$ .

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A parallel beam of monochromatic light of wavelength 5000 A is incident on a single narrow slit of width 0.001 mm. The light is focussed by a convex lens on a screen placed on focal plane. The first minimum will be formed formed for the angle of diffraction equal to

The correct option is A 20∘dsinθ=mλθ=sin−1(mλd)=sin−1((1)(5000×10−10)0.001×10−3)θ=30o.

The correct option is A $20^{\circ}$
$d sin θ = m λ$
$θ = {sin}^{- 1} (\frac{m λ}{d})$
$= {sin}^{- 1} (\frac{(1) (5000 \times 10^{- 10})}{0.001 \times 10^{- 3}})$
$θ = 30^{o}$ .