Angular width (beta) of the central maximum o | vedclass

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

Question

(A) For a single slit diffraction pattern, the condition for the first minimum is $d \sin 	heta = \lambda$, where $d$ is the slit width and $\lambda$

Vedclass · Accepted Answer

Distance between the slit and the screen

Vedclass · Answer

(A) For a single slit diffraction pattern, the condition for the first minimum is $d \sin 	heta = \lambda$, where $d$ is the slit width and $\lambda$ is the wavelength of light.Since $	heta$ is very small, $\sin 	heta \approx 	heta$, so $	heta = \frac{\lambda}{d}$.The angular width of the central maximum is $2	heta = \frac{2\lambda}{d}$.This expression shows that the angular width depends on the wavelength $(\lambda)$ and the slit width $(d)$.Since frequency $(f)$ is related to wavelength by $\lambda = \frac{c}{f}$, the angular width also depends on the frequency.However, it does not depend on the distance $(D)$ between the slit and the screen.

Angular width $(\beta)$ of the central maximum of a diffraction pattern on a single slit does not depend upon

Similar Questions

Describe the simplest experiments to observe a diffraction pattern.

In Fraunhofer diffraction by a single slit,if the slit width is $a$,the wavelength is $\lambda$,and the focal length of the lens is $f$,what is the linear width of the central maximum?

$A$ circular disc is placed in front of a narrow source. When the point of observation is at a distance of $1 \, m$ from the disc,the disc covers the first $HPZ$ (Half Period Zone). The intensity at this point is $I_0$. The intensity at a point at a distance of $25 \, cm$ from the disc will be:

Consider sunlight incident on a slit of width $10^4 \ \mathring{A}$. The image seen through the slit shall

Vedclass Test Series

Exam Paper Generator

Online Exam Module