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Light of wave length λ is incident on a slit. First mimima of the diffraction pattern is found to lie at a distance of 6mm from the central maximum on a screen placed at a distance of 2m from the slit. If slit width is 0.2mm then the wavelength of the light used will be

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$7400 A^{0}$

$7000 A^{0}$

$4000 A^{0}$

$6000 A^{0}$

answer is B.

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Detailed Solution

$For first maxima, y = \frac{λD}{a}$

$\Rightarrow 6 \times 10^{- 3} = \frac{λ \times 2}{0.1 \times 10^{- 3}}$

$\Rightarrow λ = 6000 \overset{0}{A}$

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