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In YDSE, the two slits are separated by 0.1 mm and they are 0.5 m from the screen. The wavelength of light used is 5000 Å. Find the distance between 7th maxima and 11 th mimima on the upper side of screen.

Answer» Given, `d = 0.1 mm =10^-4m, D = 0.5 m and lambda = 5000 Å = 5.0 xx (10^-7)m.`
`Delta y = y_11 -y_7 = ((2xx 11xx -1)lambda D)/(2d) -(7lambdaD)/(d)`
` or Delta y = (7lambdaD)/(2d) = ( 7xx 5.0 xx 10^(-7) xx 0.5)/(2 x 10^(-4))`
` = 8.75 xx (10^-3) `
` = 8.75 mm.`


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