Question

A narrow monochromatic beam of light of intensity I is incident on a glass plate. Another identical glass plate is kept 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 approximately.

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

Answer 1

The correct option is C

49 : 1

Let I = 100

Intensity of light after reflecting from 1st plate

$I_1=25$

Amplitude $A_1=\sqrt{I_1}=$5 units

After reflecting from $2^{nd}$ surface intensity $=\frac{75\times 25}{100}=\frac{75}{4}$

75% of this light pass through 1 after reflection.

$\therefore I_2=\frac{75}{4}\times \frac{75}{100}=\frac{225}{16}$

$A_2=\sqrt{\frac{225}{16}}=\frac{15}{4}=3.75$

$\frac{I_{max}}{I_{min}}={\left(\frac{5+3.75}{5-3.75}\right)}^2=49:1$