A fringe width of 6 mm was produced for two | vedclass

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

Question

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.Here,$\beta = 6 \ mm = 6 	imes 10^{-3} \

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.Here,$\beta = 6 \ mm = 6 	imes 10^{-3} \ m$,$d = 1 \ mm = 1 	imes 10^{-3} \ m$,and $D = 10 \ m$.Rearranging the formula to solve for wavelength $\lambda$:$\lambda = \frac{\beta d}{D} = \frac{(6 	imes 10^{-3} \ m) 	imes (1 	imes 10^{-3} \ m)}{10 \ m}$.$\lambda = \frac{6 	imes 10^{-6}}{10} \ m = 6 	imes 10^{-7} \ m$.To convert this to nanometers $(nm)$,we multiply by $10^9$:$\lambda = 6 	imes 10^{-7} 	imes 10^9 \ nm = 600 \ nm$.Thus,the value of $x$ is $600$.

$A$ fringe width of $6 \ mm$ was produced for two slits separated by $1 \ mm$ apart. The screen is placed $10 \ m$ away. The wavelength of light used is $'x' \ nm$. The value of $'x'$ to the nearest integer is

Similar Questions

In a Young's double-slit experiment,interference is obtained using electron waves. The electrons are produced by an electron gun. To increase the fringe width,one must:

Young's experiment demonstrates that . . . . . .

In Young's double-slit experiment using monochromatic light,the shape of the interference pattern on the screen is .....

The distance between two coherent sources is $0.1 \, mm$. The fringe-width on a screen $1.2 \, m$ away from the source is $6.0 \, mm$. The wavelength of light used is ......... $\mathring{A}$.

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module