In Young's double-slit arrangement,the screen | vedclass

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

Question

(B) Let $d$ be the distance between the slits and $\lambda$ be the wavelength of light used.At $t=0$,the distance between the slits and the screen is

Vedclass · Accepted Answer

$\frac{x}{v}$

Vedclass · Answer

(B) Let $d$ be the distance between the slits and $\lambda$ be the wavelength of light used.At $t=0$,the distance between the slits and the screen is $D = x$. The position of the $1^{	ext{st}}$ order maxima at point $A$ is given by $y = \frac{\lambda D}{d} = \frac{\lambda x}{d}$.At time $t$,the distance between the slits and the screen becomes $D_t = x + vt$.The position of the $1^{	ext{st}}$ order minima at point $A$ is given by $y = \frac{\lambda D_t}{2d} = \frac{\lambda (x + vt)}{2d}$.Since the point $A$ remains at the same position $y$,we equate the two expressions:$\frac{\lambda x}{d} = \frac{\lambda (x + vt)}{2d}$$2x = x + vt$$x = vt$$t = \frac{x}{v}$

Similar Questions

In the setup shown in the figure,the two slits,$S_1$ and $S_2$,are not equidistant from the slit $S$. The central fringe at $O$ is then

In a Young's double slit experimental arrangement shown here,if a mica sheet of thickness $t$ and refractive index $\mu$ is placed in front of the slit $S_1$,then the path difference $(S_1P - S_2P)$

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 $: 1$)?

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