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A parallel beam of light of wavelength 500nm falls on a narrow slit and the resulting diffraction pattern is observed on screen 1m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit. |
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Answer» SOLUTION :`theta=(y)/(D), theta= (2.5xx10^(-3))/(1)` radian. Now, `d sin theta = N lambda` Since `theta` is very small, therefore `sin theta = theta`. or `d= (n lambda)/(theta)=(1xx500xx10^(-9))/(2.5xx10^(-3))m = 2xx 10^(-4)m= 0.2 mm`. |
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