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A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a 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. |
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Answer» SOLUTION :Here wavelength of light `l=500nm=5xx10^(-7)m`, Distance of screen from the slit `D=1m` and distance of FIRST minimum on the screen from the central maxima `x_(1)=2.5mm=2.5xx10^(-3)m` `therefore` Diffraction ANGLE for first minima `theta_(1)=(x_(1))/(D)=(2.5xx10^(-3))/(1)=2.5xx10^(-3)rad` We know that for first diffraction minima `a sintheta_(1)=lamdaimpliesa=(lamda)/(sintheta_(1))=(lamda)/(theta_(1))` `therefore` Slit width `a=(lamda)/(theta_(1))=(5xx10^(-7))/(2.5xx10^(-3))=2xx10^(-4)m=0.2mm`. |
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