1.

In a YDSE experiment, d = 1mm, `lambda`= 6000Å and D= 1m. The minimum distance between two points on screen having 75% intensity of the maximum intensity will beA. 0.50 mmB. 0.40 mmC. 0.30 mmD. 0.20 mm

Answer» Correct Answer - D
`3/4 I_(max) = I_(max) cos^2 (phi/2)`
` :. phi/2 = pi/12` and `(5pi)/6`
`:. phi= pi/6`
and `(5pi)/3 = ((2pi)/lambda) (Deltax) = ((2pi)/lambda) (yd)/D`
`:. y_1= (lambdaD)/(12d) = ((6000 xx 10^-10)(1))/((12)(10^-3))`
`= 0.05 xx (10^-3) m = 0.05mm`
`y_2 =5 ((lambdaD)/(12d))= 5xx 0.05 mm `
=0.25mm
`Deltay = y_2-y_1 = 0.2mm` .


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