In a Young's double slit experiment,each of t | vedclass

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

Question

(A) $(1)$ The position of the $n^{	ext{th}}$ bright fringe is given by $y = n \frac{\lambda D}{d}$.For the $8^{	ext{th}}$ fringe,$y = 8 \frac{\lambd

Vedclass · Accepted Answer

$601.50, 24$

Vedclass · Answer

(A) $(1)$ The position of the $n^{	ext{th}}$ bright fringe is given by $y = n \frac{\lambda D}{d}$.For the $8^{	ext{th}}$ fringe,$y = 8 \frac{\lambda D}{d}$.The extreme positions are $y_{\max} = 8 \frac{\lambda D}{d_{\min}}$ and $y_{\min} = 8 \frac{\lambda D}{d_{\max}}$.The separation is $\Delta y = y_{\max} - y_{\min} = 8 \lambda D \left[ \frac{1}{d_{\min}} - \frac{1}{d_{\max}} ight]$.Given $\lambda = 6000 	imes 10^{-10} \ m$,$D = 1 \ m$,$d_{\max} = 0.84 \ mm = 0.84 	imes 10^{-3} \ m$,and $d_{\min} = 0.76 \ mm = 0.76 	imes 10^{-3} \ m$.$\Delta y = 8 	imes 6000 	imes 10^{-10} 	imes 1 	imes \left[ \frac{1}{0.76 	imes 10^{-3}} - \frac{1}{0.84 	imes 10^{-3}} ight] = 48 	imes 10^{-7} 	imes \left[ \frac{0.08}{0.76 	imes 0.84 	imes 10^{-3}} ight] \approx 601.5 \ \mu m$.$(2)$ The speed is $v = \frac{dy}{dt} = \frac{d}{dt} \left( n \frac{\lambda D}{d} ight) = -n \lambda D \frac{1}{d^2} \frac{dd}{dt}$.Given $d = 0.8 + 0.04 \sin \omega t$,so $\frac{dd}{dt} = 0.04 \omega \cos \omega t$.For maximum speed,$\cos \omega t = 1$,so $\frac{dd}{dt} = 0.04 \omega = 0.04 	imes 0.08 = 0.0032 \ mm/s$ and $d = 0.8 \ mm$.$v_{\max} = \left| -8 	imes 6000 	imes 10^{-10} 	imes 1 	imes \frac{0.0032 	imes 10^{-3}}{(0.8 	imes 10^{-3})^2} ight| = 24 \ \mu m/s$.

Similar Questions

Each plate reflects $25\%$ of the incident light intensity. When $AB$ and $A'B'$ are taken as the two slits in Young's double-slit experiment,what is the ratio of maximum to minimum intensity ${I_{\max }}/{I_{\min }}$ (in $: 1$)?

$A$ transparent film $(\mu=1.45)$ of thickness $0.02 \ mm$ is placed on one of the slits of a Young's double slit experiment which uses monochromatic light of wavelength $620 \ nm$. How many fringes will cross through the center if the film is removed?

In Young's double slit experiment,a glass plate is placed before one slit which absorbs half the intensity of light. Under this case:

$A$ parallel beam of light of wavelength $500 \ nm$ is incident at an angle $30^o$ with the normal to the slit plane in a Young's double-slit experiment. The intensity due to each slit is $I_o$. Point $O$ is equidistant from $S_1$ and $S_2$. The distance between the slits is $1 \ mm$.

If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module