Question

A narrow monochromatic beam of light of intensity I is incident on a glass plate as shown in fig. Another identical glass plate is kept close to the first one and parallel to it. Each glass plate reflects 25% of the light incident on it and transmits the remaining. Then the ratio of the maximum to minimum intensities in the interference pattern formed by the two beams obtained after one reflection at each plate is –

• A. 7 : 1
• B. 49 : 1
• C. 4 : 1
• D. 16 : 9

Answer 1

Answer: (B)

49 : 1

Answer 2

Intensity of light after reflection through I^st plate,

I₁ = $\frac{I}{4}$. Intensity of light after reflection through 2 & finally transmission through 2^nd plate. I₂ = $\frac{aI}{64}$

$\frac{I_{\max}}{I_{\min}}$=$\left\lbrack \frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}} \right\rbrack^{2}$ = 49 : 1