In a single slit diffraction pattern, the distance between first minima on the right and first minima on the left of central maximum is 4 mm. The screen on which the pattern is displaced, is 2m from the slit and wavelength of light used is 6000Å. Calculate width of the slit and width of the central maximum.

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#### Solution

Given:- λ = 6000 Å = 6 x 10^{-7} m,

D = 2 m,

X_{1} + X_{2} = 4 mm = 4 x 10^{-3} m

To find:- Width of slit (a)

Width of central maximum (W)

Formula:-

i. `X_1+X_2=(2lambdaD)/a`

ii. W = X_{1} + X_{2}

Calculation: From formula (i),

`4xx10^-3=(2xx6xx10^-7xx2)/a`

`therefore a=6xx10^4m`

**The width of the slit is 6 x 10 ^{-4} m.**

From formula (ii),

Width of central maximum = 4 x 10^{-3} m

**The width of the central maximum is 4 x 10 ^{-3} m.**

Concept: Fraunhofer Diffraction Due to a Single Slit

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