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Question
A parallel beam of light of 450 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.5 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit.
Solution
The distance of the nth minimum from the center of the screen is, `x_n =(nDlambda)/a`
Where, D = distance of slit from screen, λ = wavelength of the light, a = width of the slit
For first minimum, n = 1:
`3 xx 10^-3 = (1xx (1.5) xx (450 xx 10^-9))/a => a = (1 xx (1.5) xx (450 xx 10^-9))/(3 xx 10^-3)`
∴ a = 0.225 mm
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