A parallel beam of light of intensity `I_@` is incident on a coated glass plate. If 25% of the incident light is reflected from the upper surface and 50% of light is reflected from the lower surface of the glass plate, the ratio of maximum to minimum intensity in the interference region of the reflected light is

Options

A. `((1/2+sqrt(3/8))/(1/2 - sqrt(3/8)))^2`
B. `((1/4+sqrt(3/8))/(1/2 - sqrt(3/8)))^2`
C. 5/8
D. 8/5

Answer

B

Explanation

From the above figure, .
`I_1 = I_@/4` .
` I_2 = 3/8 I_@`.
Maximum to minimum intensity is,
`I_max / I_min = [(sqrtI_1 + sqrtI_2)^2]/[(sqrtI_1 - sqrtI_2)^2] = [[sqrt(I_@/4) + sqrt(3I_@/8)]]^2/ [[sqrt(I_@/4) - sqrt(3I_@/8)]]^2 `.
`[I_@[sqrt(I/4) + sqrt(3/8)]]^2/ [I_@[sqrt(1/4) - sqrt(3/8)]]^2` .
` therefore I_max/I_min= [(1/2 + sqrt(3/8)]/ [1/2 - sqrt(3/8))]^2 `