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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is obtained on a screen 1 m away. If the first minimum is formed at a distance of 2.5 mm from the centre of the screen, find the (i) width of the slit, and (ii) distance of first secondary maximum from the centre of the screen. |
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Answer» Solution :Here wavelength of light waves `LAMBDA=500 nm=500xx10^(-9)m`, distance of screen from the slit D = 1m, distance of first minima from the centre of screen `x_(1)=2.5m m=2.5xx10^(-3)m`. (i) `because x_(1)=D theta_(1)= (D lambda)/(a)`, where a = slit width `RARR a=(Dlambda)/(x_(1))=(1xx500xx10^(-9))/(2.5xx10^(-3))=2xx10^(-4)m or 0.2`MM (ii) The distance of first secondary maxima from the centre of the screen `x._(1) =(3)/(2)(Dlambda)/(a)=(3)/(2) x_(1)=(3)/(2) XX 2.5xx10^(-3)=3.75xx10^(-3)m` or 3.75 mm |
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