JEE MAIN - Physics (2023 - 1st February Evening Shift - No. 21)
As shown in the figure, in Young's double slit experiment, a thin plate of thickness $$t=10 \mu \mathrm{m}$$ and refractive index $$\mu=1.2$$ is inserted infront of slit $$S_{1}$$. The experiment is conducted in air $$(\mu=1)$$ and uses a monochromatic light of wavelength $$\lambda=500 \mathrm{~nm}$$. Due to the insertion of the plate, central maxima is shifted by a distance of $$x \beta_{0} . \beta_{0}$$ is the fringe-width befor the insertion of the plate. The value of the $$x$$ is _____________.
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Answer
4
Explanation
Given $\mathrm{t}=10 \times 10^{-6} \mathrm{~m}$
$$ \begin{aligned} & \mu=1.2 \\\\ & \lambda=500 \times 10^{-9} \mathrm{~m} \end{aligned} $$
When the glass slab inserted infront of one slit then the shift of central fringe is obtained by
$ \mathrm{t}=\frac{\mathrm{n} \lambda}{(\mu-1)} $
$ \Rightarrow \quad 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{(1.2-1)} $
$$ \Rightarrow $$ $ 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{0.2} $
$$ \Rightarrow $$ $ \mathrm{n}=4 $
$$ \begin{aligned} & \mu=1.2 \\\\ & \lambda=500 \times 10^{-9} \mathrm{~m} \end{aligned} $$
When the glass slab inserted infront of one slit then the shift of central fringe is obtained by
$ \mathrm{t}=\frac{\mathrm{n} \lambda}{(\mu-1)} $
$ \Rightarrow \quad 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{(1.2-1)} $
$$ \Rightarrow $$ $ 10 \times 10^{-6}=\frac{\mathrm{n} \times 500 \times 10^{-9}}{0.2} $
$$ \Rightarrow $$ $ \mathrm{n}=4 $
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