The central fringe of the interference patter | vedclass

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

Question

(A) The shift in the interference pattern due to the introduction of a glass plate is given by the formula: $\Delta x = \frac{t(\mu - 1)D}{d}$.Given t

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(A) The shift in the interference pattern due to the introduction of a glass plate is given by the formula: $\Delta x = \frac{t(\mu - 1)D}{d}$.Given that the central fringe shifts to the position of the $4^{th}$ bright fringe,the shift is equal to $4 \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.Equating the shift: $\frac{t(\mu - 1)D}{d} = 4 \frac{\lambda D}{d}$.This simplifies to: $t(\mu - 1) = 4 \lambda$.Given $\lambda = 6000 \, \mathring{A} = 6 	imes 10^{-7} \, m$ and $\mu = 1.5$,we have:$t(1.5 - 1) = 4 	imes 6 	imes 10^{-7} \, m$.$t(0.5) = 24 	imes 10^{-7} \, m$.$t = \frac{24 	imes 10^{-7}}{0.5} = 48 	imes 10^{-7} \, m$.$t = 4.8 	imes 10^{-6} \, m = 4.8 \, \mu m$.

The central fringe of the interference pattern produced by light of wavelength $6000 \, \mathring{A}$ is found to shift to the position of the $4^{th}$ bright fringe after a glass plate of refractive index $1.5$ is introduced. The thickness of the glass plate would be ........ $\mu m$.

Similar Questions

Which of the following substances would quickly absorb oxygen?

The $\Delta H^o$ of combustion of $C_4H_{10(g)}$ is $-2650 \ kJ \ mol^{-1}$. The amount of heat evolved at constant pressure that can be obtained by burning $5.8 \ kg$ of $C_4H_{10}$ is approximately:

The potential energy between two atoms in a molecule is given by $U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}$,where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atoms are in stable equilibrium when:

If $x > 0, y > 0, z > 0, xy + yz + zx < 1$ and if $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$,then $x + y + z$ equals to

For a first order reaction,the concentration of reactant was reduced from $0.03 \ mol \ L^{-1}$ to $0.02 \ mol \ L^{-1}$ in $25 \ min$. What is its rate (in $mol \ L^{-1} \ s^{-1}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module