A parallel beam of light of intensity I_0 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident li

Vedclass · Accepted Answer

$\left[\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}\right]^2$

Vedclass · Answer

(A) Given that $25 \%$ of the total intensity of incident light is reflected from the upper surface. This implies that if the intensity of incident light is $I_0$,the intensity of the first reflected ray is $I_1 = 0.25 I_0 = \frac{I_0}{4}$.The intensity of light reaching the lower surface of the plate is $I_0 - 0.25 I_0 = 0.75 I_0 = \frac{3}{4} I_0$.Since $50 \%$ of this intensity is reflected from the lower surface,the intensity of the second reflected ray is $I_2 = 0.50 	imes \frac{3}{4} I_0 = \frac{3}{8} I_0$.The ratio of maximum to minimum intensity is given by the formula:$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{I_1}+\sqrt{I_2}ight)^2}{\left(\sqrt{I_1}-\sqrt{I_2}ight)^2}$Substituting the values of $I_1$ and $I_2$:$\frac{I_{\max }}{I_{\min }}=\frac{\left(\sqrt{\frac{I_0}{4}}+\sqrt{\frac{3 I_0}{8}}ight)^2}{\left(\sqrt{\frac{I_0}{4}}-\sqrt{\frac{3 I_0}{8}}ight)^2} = \left(\frac{\frac{1}{2} \sqrt{I_0} + \sqrt{\frac{3}{8}} \sqrt{I_0}}{\frac{1}{2} \sqrt{I_0} - \sqrt{\frac{3}{8}} \sqrt{I_0}}ight)^2 = \left(\frac{\frac{1}{2}+\sqrt{\frac{3}{8}}}{\frac{1}{2}-\sqrt{\frac{3}{8}}}ight)^2$

$A$ parallel beam of light of intensity $I_0$ is incident on a glass plate. $25 \%$ of light is reflected by the upper surface and $50 \%$ of light is reflected from the lower surface. The ratio of maximum to minimum intensity in the interference region of the reflected rays is

Similar Questions

In a biprism experiment,if the $5^{\text{th}}$ bright band with wavelength $\lambda_1$ coincides with the $6^{\text{th}}$ dark band with wavelength $\lambda_2$,then the ratio $(\lambda_1 / \lambda_2)$ is:

White light is incident at an angle of $30^\circ$ on a soap film with a refractive index of $4/3$. The wavelength of the transmitted light is observed to be $6 \times 10^{-5} \, cm$. Find the minimum thickness of the film.

In a biprism experiment,the fringe width is $0.6 \ mm$. The distance between the $6^{th}$ dark fringe and the $8^{th}$ bright fringe on the same side of the central bright fringe is: (in $mm$)

$A$ thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown. The observed interference fringes from this combination shall be

$A$ mixture of yellow light of wavelength $580 \ nm$ and blue light of wavelength $450 \ nm$ is incident normally on an air film of thickness $2.9 \times 10^{-4} \ mm$. The colour of reflected light is

Vedclass Test Series

Exam Paper Generator

Online Exam Module