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In a single slit diffraction the distance between the first minima on either side of the central maximum is 5mm. The screen is place data distance of 75 cm from the slit and the wavelength of light used is 546nm. Calculate the slit width. |
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Answer» Solution :Distance between FIRST MINIMUM and central maximum`y=5/2=2.5mm=2.5xx10^(-3)m` We have `THETA=(n lamda)/a, n=1`and `lamda=546xx10^(-9)m` Also `theta=Y/D, D=75cm=75xx10^(-2)m` `:.(n lamda)/a=y/d` `a=(nlamdaD)/y=(75xx10^(-2)xx1xx546xx10^(-9))/(2.5xx10^(-3))` i.e. slit WIDTH`a=1.638xx10^(-4)m=0.1638mm` |
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