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Light of wavelength 6000 $\overset{o}{A}$ is incident on a single slit. First minimum is obtained at a distance of 4 mm from the centre. If the width of the slit is 0.3 mm, then distance between slit and screen will be

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1 m

1.5 m

2 m

2.5 m

answer is C.

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Detailed Solution

Let distance of first minimum be $y .$

$∴ a sinθ = 1 \cdot λ \Rightarrow a \cdot \frac{y}{D} = λ \Rightarrow D = \frac{ay}{λ} = \frac{0.3 \times 4 \times 10^{- 6}}{6000 \times 10^{- 10}} m$

$\Rightarrow D = 2 m$

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