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  • In Young's double slit experiment, two slits S1 and S2 are ' d ' distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal

In Young's double slit experiment, two slits S1 and S2 are ' d ' distance apart and the separation from slits to screen is D (as shown in figure). Now if two transparent slabs of equal thickness 0.1 mm but refractive index 1.51 and 1.55 are introduced in the path of beam (\lambda=4000 \AA) from S_{1}and S_{2} respectively. The central bright fringe spot will shift by number of fringes. 

Option: 1

10


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

\begin{array}{ll} \mu_1=1.51 & t=0.1 \mathrm{~mm} \\ \mu_2=1.55 & \lambda=4000 \AA \end{array}

Shifting central maxima

\begin{aligned} & \Delta x=\left[S_1 P+\left(\mu_1-1\right) t\right]-\left[S_2 P+\left(\mu_2-1\right) t\right] \\ & 0=\left(S_1 P-S_2 P\right)+\left(\mu_1-1\right) t-\left(\mu_2-1\right) t \\ & 0=\frac{y d}{D}+\left(\mu_1-\mu_2\right) t \\ & \left(\mu_2-\mu_1\right) t=\frac{y d}{D} \end{aligned}

\begin{aligned} & (1.55-1.51)(0.1 \mathrm{~mm})=y \times \frac{d}{D} \\ & \frac{D}{d}(0.04 \times 0.1) \times 10^{-3}=y \end{aligned}

\text { Fringe width } \Rightarrow \beta=\frac{\lambda D}{d}

\text { No. of fringes shifted }=\frac{y}{\beta}=\frac{4 \times 10^{-6}}{4000 \AA}=10

 

Posted by

HARSH KANKARIA

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