A screen is placed at a distance 50,cm from a | vedclass

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

Question

(B) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.Giv

Vedclass · Accepted Answer

$0.230$

Vedclass · Answer

(B) The position of the $n^{th}$ minima in a single-slit diffraction pattern is given by $x_n = \frac{n D \lambda}{a}$,where $a$ is the slit width.Given: $D = 50\,cm = 0.5\,m$,$\lambda = 690\,nm = 690 	imes 10^{-9}\,m$,and the distance between the first $(n=1)$ and third $(n=3)$ minima is $\Delta x = x_3 - x_1 = 3.00\,mm = 3 	imes 10^{-3}\,m$.Using the formula: $\Delta x = (3 - 1) \frac{D \lambda}{a} = \frac{2 D \lambda}{a}$.Rearranging for $a$: $a = \frac{2 D \lambda}{\Delta x}$.Substituting the values: $a = \frac{2 	imes 0.5 	imes 690 	imes 10^{-9}}{3 	imes 10^{-3}} = \frac{690 	imes 10^{-9}}{3 	imes 10^{-3}} = 230 	imes 10^{-6}\,m = 0.230\,mm$.

$A$ screen is placed at a distance $50\,cm$ from a single slit,which is illuminated with light of wavelength $690\,nm$. If the distance between the first and third minima is $3.00\,mm$,what is the width of the slit (in $,mm$)?

Similar Questions

$A$ plane wave front of wavelength $\lambda$ is incident on a single slit of width $b$. What is the angular width of the secondary maximum?

Why does a $DVD$ store almost $30$ times more information than a $CD$?

In a single slit diffraction pattern,the intensity and width of the fringes are:

When you look at a clear blue sky, you see tiny specks and hair-like structures floating in your view, called "floaters". This is basically

Angular width of the central maximum of a diffraction pattern due to a single slit does not depend upon:

Vedclass Test Series

Exam Paper Generator

Online Exam Module