JEE Class main Answered
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Asked by eknathmote50 | 24 May, 2019, 11:47: AM
Expert Answer
Intensity I = I1 + I2 + 2 (I1 I2 )1/2cosδ
where I1 and I2 are intensities of light coming out of slit-1 and slit-2
If I1 = I and I = I , then by using cosδ=+1, we have maximum intensity Imax = 4I
by using cosδ=-1, we have minimum intensity Imin = 0
If I1 = I and I = I/2 , then by using cosδ=+1, we have maximum intensity Imax = I [ (3/2) + (2/√2) ] = I[ (3/2) + √2 ]
by using cosδ=-1, we have minimum intensity Imin = I [ (3/2) - (2/√2) ] = I[ (3/2) - √2 ]
Hence if intensity of one slit is halved by introducing a transparent plate, intensity of bright fringe is getting reduced
and intensity of dark fringe is increased
Answered by Thiyagarajan K | 25 May, 2019, 06:47: AM
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