A parallel beam of monochromatic light of intensity I0 is incident on a glass plate, 25% of light is reflected by upper surface and 50% of light is reflected from lower surface. The ratio of minimum and maximum intensity in interference region of reflected light is
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a
12−38212+382
b
1−221+22
c
38
d
49
answer is A.
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Detailed Solution
Intensity of reflected light from upper surface is I1=I0×25100=I04 And intensity of transmitted light from upper surface is I'=I0−I04=3I04 The intensity of reflected light from lower surface is I2=3I04×50100=3I08 ∴IminImax=I1−I22I1+I22=I04−3I082I04+3I082 = 12−38212+382