Question

A parallel beam of light of intensity $I_0$ 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

• A. ${\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)}^2$
• B. ${{\left(\frac{\frac{1}{4}+\sqrt{\frac{3}{8}}}{\frac{1}{4}-\sqrt{\frac{3}{8}}}\right)}^2}^{}$
• C. $\frac{5}{8}$
• D. $\frac{8}{5}$

Answer 1

Answer: (A)

${\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)}^2$

Intensity of reflected light from the upper surface $I_1=\frac{I_0}{4}$
Intensity of transmitted light through the upper surface $I_2=\frac{3}{4}{I}_0$
Intensity of reflected light from the lower surface $I_3=\frac{I_2}{2}=\frac{3I_o}{8}$
Maximum intensity $\frac{I_{max}}{I_{min}}=\frac{{(\sqrt{I_1}+\sqrt{I_3})}^2}{{(\sqrt{I_1}-\sqrt{I_3})}^2}$
Or $I_{max}={\left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right)}^2$