In a single -slit diffraction experiment the slit width is 0 . 12 mm and the wavelength of the light used is 6 . 00 × 10 - 7 m . Calculate the width of the central maxima on a screen 2 . 00 m from the slit.

Given, The slit-width, d = 0 . 12 mm The wavelength of the light, λ = 6 . 00 × 10 - 7 m The distance between the slit and the screen, D = 2 . 00 m We know that, d sin θ = λ sin θ ≈ θ = x D = λ d x D = λ d x 2 . 00 = 6 . 00 × 10 - 7 0 . 12 × 10 - 3 x = 6 . 00 × 10 - 7 0 . 12 × 10 - 3 × 2 . 00 x = 0 . 01 m Now, the width of the central maxima on the screen is 2 x = 2 × 0 . 01 = 0 . 02 m .

The intensity pattern for single slit diffraction is shown in the diagram (The vertical units are arbitrary) The wavelength of the light used is λ b On a copy of the axis draw a graph to show how the intensity varies with diffraction angle for a slit with (i) Width b 2 and wavelength λ

In single-slit diffraction, intensity is given by the equation, I = I 0 sin β β 2 where β = πbsinθ λ b = slit width The intensity verse angle of diffraction when the slit width is halved. From the above graph, we can conclude that when the slit width is halved. Hence, the diffraction pattern will be wider.

The intensity pattern for single slit diffraction is shown in the diagram (The vertical units are arbitrary) The wavelength of the light used is λ b On a copy of the axis draw a graph to show how the intensity varies with diffraction angle for a slit with (i) Width b 2 and wavelength λ / 2

In single-slit diffraction, intensity is given by the equation, I = I 0 sin β β 2 where β = πbsinθ λ b = slit width The intensity verse angle of diffraction when the slit width is halved and wavelength of the light is also halved. From the above graph, we can conclude that when the slit width is halved and wavelength is also halved. Hence, with no effect on the intensity by decreasing in wavelengths, the diffraction pattern will be wider.