In a single slit diffraction, the width of slit 0.5 cm, focal length of lens is 40 cm and wavelength of light is 4890Å. The distance of first dark fringe from central maxima is

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40µm

20µm

60µm

80µm

answer is B.

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Detailed Solution

$\begin{array}{l} asin θ = nλ; a \frac{y}{d} = nλ \\ \frac{0 .5 \times 10^{- 2} y}{40 \times 10^{- 2}} = (1) (4890 \times 10^{- 10}) \\ y = \frac{4890 \times 10^{- 10} \times 40}{0 .5} = m \\ = 39 .1 \times 10^{- 6} m ≃ 40 μm \end{array}$

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