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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observe on screen 1 m 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.A. 0.2 mmB. 0.3 mmC. 0.4 mmD. 0.5 mm |
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Answer» Correct Answer - A Distance of first minima in diffraction pattern due to a single slit is `x_(1)=(Dlambda)/(a)` It is given that `D=1m, lambda=500 mm=500xx10^(-19)m` `x_(1)=2.5mm=2.5 xx 10^(-3)m, a=?` `therefore a=(Dlambda)/(x_(1))=(1xx500xx10^(-9))/(2.5xx10^(-3)) m ` `=200xx10^(-6)m=0.2 mm` |
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