Two identical glass plates are kept close and parallel to each other at one of the slit in YDSE arrangement - Physics - Wave Optics

Question

Two identical glass plates are kept close and parallel to each
other at one of the slit in YDSE arrangement Each glass plate
reflects 20% of the light incident on it and rest is transmitted If
slits are illuminated with light of same intensity then ratio of
the minimum and maximum intensities in the interference formed is

(A) 1/81 (B) 1/9 (C) 1/4 (D) 1/16

Jyoti Pant · Accepted Answer

Suppose I is the intensity of incident beam
According to question, 20% i e
 1/5th of incident light is reflected
Thus, the intensity of beam reflected from first plate=I1=I5So, the intensity of light incident on second plate will be I-I5=4I5And as 20% of it is again reflected from second plate, so intensity of reflected beam is15×4I5=4I25Intensity of transmitted light, I2=[1-15]×4I25=45×4I25=16I125For interference,IR=I1+I2+2I1I2cosϕIR=max if cosϕ=max=1So, IR=I1+I2+2I1I2=[I1+I2]2=[I/5+16I/125]2=I5+16I125+216I2625=I5+16I125+8I25=25I+16I+40I125=81I125Again, IR=min if cosϕ=min=-1 IR=I1+I2-2I1I2=[I1-I2]2=[I/5-16I/125]2=I5+16I125-216I2625=I5+16I125-8I25=25I+16I-40I125=I125So,ImaxImin=81I125×125I=811Or Imin:Imax=1:81Thus, the correct option is (A)