Two slits in Young's experiment have widths i | vedclass

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

Question

(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.$	herefo

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(B) The intensity $I$ of light is directly proportional to the width $W$ of the slit,and also to the square of the amplitude $A$ of the wave.$	herefore \frac{I_1}{I_2} = \frac{W_1}{W_2} = \frac{A_1^2}{A_2^2}$Given the ratio of widths $\frac{W_1}{W_2} = \frac{1}{25}$,we have:$\frac{A_1^2}{A_2^2} = \frac{1}{25} \implies \frac{A_1}{A_2} = \sqrt{\frac{1}{25}} = \frac{1}{5}$The ratio of maximum intensity to minimum intensity is given by:$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2} = \left( \frac{\frac{A_1}{A_2} + 1}{\frac{A_1}{A_2} - 1} ight)^2$Substituting $\frac{A_1}{A_2} = \frac{1}{5}$:$\frac{I_{max}}{I_{min}} = \left( \frac{\frac{1}{5} + 1}{\frac{1}{5} - 1} ight)^2 = \left( \frac{\frac{6}{5}}{-\frac{4}{5}} ight)^2 = \left( -\frac{6}{4} ight)^2 = \left( -\frac{3}{2} ight)^2 = \frac{9}{4}$

Two slits in Young's experiment have widths in the ratio $1 : 25$. The ratio of intensity at the maxima and minima in the interference pattern,$\frac{I_{max}}{I_{min}}$ is

Similar Questions

The intensity ratio of the maxima and minima in an interference pattern produced by two coherent sources of light is $9: 1$. The intensities of the light sources used are in the ratio (in $: 1$)

The ratio of intensities of two interfering waves is $9:1$. The ratio of the maximum amplitude to the minimum amplitude of the resultant wave is ........

In a Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of light used) is $I$. If $I_0$ denotes the maximum intensity,then $\frac{I}{I_0} = $ . . . . . .

The ratio of the intensity at the centre of a bright fringe to the intensity at a point one-quarter of the fringe width from the centre is

In Young's double slit experiment,if the maximum intensity is $I$,then the angular position where the intensity becomes $\frac{I}{4}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module