In Young's double slit experiment,the intensi | vedclass

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

Question

(A) Given that the intensity of the first slit $I_1 = 2I_2$,where $I_2$ is the intensity of the second slit.The ratio of maximum intensity to minimum

Vedclass · Accepted Answer

$34$

Vedclass · Answer

(A) Given that the intensity of the first slit $I_1 = 2I_2$,where $I_2$ is the intensity of the second slit.The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{I_1} + \sqrt{I_2}}{\sqrt{I_1} - \sqrt{I_2}} \right)^2$Substituting $I_1 = 2I_2$ into the formula:$\frac{I_{\max}}{I_{\min}} = \left( \frac{\sqrt{2I_2} + \sqrt{I_2}}{\sqrt{2I_2} - \sqrt{I_2}} \right)^2 = \left( \frac{\sqrt{2} + 1}{\sqrt{2} - 1} \right)^2$Rationalizing the denominator:$\frac{I_{\max}}{I_{\min}} = \left( \frac{(\sqrt{2} + 1)(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \right)^2 = (\sqrt{2} + 1)^4 = (2 + 1 + 2\sqrt{2})^2 = (3 + 2\sqrt{2})^2$Calculating the value:$\frac{I_{\max}}{I_{\min}} = 9 + 8 + 12\sqrt{2} = 17 + 12(1.414) \approx 17 + 16.968 \approx 33.968 \approx 34$.

In Young's double slit experiment,the intensity of light coming from the first slit is double the intensity from the second slit. The ratio of the maximum intensity to the minimum intensity on the interference fringe pattern observed is

Similar Questions

Waves emitted from two identical sources produce an intensity of $K$ units at a point on the screen. If the path difference between these two waves is $\lambda$,calculate the intensity at a point on the screen where the path difference is $\lambda/4$.

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 the light used) is $I$. If $I_0$ denotes the maximum intensity,the ratio $\frac{I_0}{I}$ is equal to:

Two coherent sources of intensity ratio $\alpha$ interfere. The value of $\frac{I_{max} - I_{min}}{I_{max} + I_{min}}$ is

Obtain the conditions for constructive interference and destructive interference.

The ratio of intensities of two coherent sources is $p$. The visibility of the fringes in the interference pattern is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module