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A parallel beam of light of wavelength 500nm falls on a narrow slit and resulting diffraction pattern is observed on a 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 :The distance of the `N^(TH)` minimum from the centre of the screen is, `x_(n)=(nDlambda)/(a)`, where, `D=` distance of SLIT from screen, `lambda=` wavelength of the LIGHT, `a=` width of the slit. `n=1` (For first minimum) Thus, `2.5xx10^(-3)=(1xx1(500xx10^(-9)))/(a)` `impliesa=2xx10^(-4)m` `=0.2mm` |
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