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Question

# Draw a graph to show the relative intensity distribution for a single slit diffraction pattern. Obtain the expression for the width of central maxima. If slit width is doubled then what is the effect on width of central maxima?

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Solution

## Let the width of the slit be through which diffraction pattern is obtained be b and λ be the wavelength of light used.The first minima of diffraction pattern is formed at point P on one side and at A of other side.Path difference between the light rays reaching at point P Δx=S2P−S1P∴ Δx=b2sinθFor first minima, path difference must be equal to λ2∴ b2sinθ=λ2OR bsinθ=λAs θ is very small, thus sinθ≈tanθ=yDOR b×yD=λ ⟹y=λDbWidth of central maxima PA=2y=2λDb⟹ PA∝1bHence the width of central maxima formed becomes half of its initial value as the slit width is doubled.

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