In a Fresnel diffraction arrangement,the scre | vedclass

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

Question

(C) The radius of the $n^{th}$ Fresnel zone is given by the formula $r_n = \sqrt{n b \lambda}$,where $b$ is the distance from the aperture to the scre

Vedclass · Accepted Answer

$5/4$

Vedclass · Answer

(C) The radius of the $n^{th}$ Fresnel zone is given by the formula $r_n = \sqrt{n b \lambda}$,where $b$ is the distance from the aperture to the screen and $\lambda$ is the wavelength of light.Given that the radius of the $4^{th}$ zone for wavelength ${\lambda _1}$ is equal to the radius of the $5^{th}$ zone for wavelength ${\lambda _2}$,we have:$r_4 = \sqrt{4 b \lambda_1}$$r_5 = \sqrt{5 b \lambda_2}$Since $r_4 = r_5$,we equate the expressions:$\sqrt{4 b \lambda_1} = \sqrt{5 b \lambda_2}$Squaring both sides,we get:$4 b \lambda_1 = 5 b \lambda_2$Dividing both sides by $b$ (where $b 
eq 0$):$4 \lambda_1 = 5 \lambda_2$Therefore,the ratio is:$\frac{\lambda_1}{\lambda_2} = \frac{5}{4}$

Similar Questions

The diagrams below show the intensity distribution in diffraction of light of two sources. In which of the following cases the sources are just resolved?

The cause of Fraunhofer lines is

What do the Fraunhofer lines in the solar spectrum indicate?

In Fresnel diffraction,if the distance between the disc and the screen is decreased,the intensity of the central bright spot will

$A$ Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance $D$ from the slit whose slit width is $a$. The distance of the first secondary maximum from the central maximum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module