Question

A parallel beam of light of 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Calculate the width of the slit. (2)

Answer 1

The distance of the nth minimum from the center of the screen is,Where, D = distance of slit from screen, λ = wavelength of the light, a = width of the slitFor first minimum, n=1Thus,

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\purple{Solution:}}}}}}$

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$\hookrightarrow$ Wavelength of the light beam, $\sf \lambda\:=\:500\,nm\:=\:500\times 10^{-9}m$

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$\hookrightarrow$ Distance of the screen from the slit, $\sf D\:=\:1m$

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$\hookrightarrow$ For first minima, $\sf n\:=\:1$

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$\hookrightarrow$ Distance between the slits = $\sf{d}$

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Distance of the first minimum from the centre of the screen can br obtained as,

$\sf x\:=\:2.5\,mm\:=\:2.5\times 10^{-3}m$

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it is related to order of minima as,

$\sf \longrightarrow\:n\lambda\:=\:x \dfrac{d}{D}$

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$\sf d\:=\: \dfrac{n\times \lambda\times D}{x}$

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$\sf d\:=\: \dfrac{1\times 500\times 10^{-9}\times 1}{2.5\times 10^{-3}}$

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$\sf d\:=\:2\times 10^{-4}m\:=\:0.2\,mm$

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$\hookrightarrow$ Therefore, the width of the slits is $\sf{\green{0.2\,mm}}$