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In a Young's double slit experiment, a monochromatic light of wavelength $600 \mathrm{~nm}$ is used. If the two slits are covered by transparent sheets of thickness $0.132 \mathrm{~mm}$ and $0.1 \mathrm{~mm}$ of refractive index 1.5 , then the number of fringes that will shift due to introduction of the sheets are
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27
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Number of fringes shift due to extra path difference caused by transparent slab is
$$
n=\frac{(\mu-1) t}{\lambda}
$$
So, when two plates of thickness $t_1$ and $t_2$ are placed infront of slits, then shift is
$$
\begin{aligned}
n & =n_2-n_1=\frac{(\mu-1)\left(t_2-t_1\right)}{\lambda} \\
& =\frac{(1.5-1)(0132-0 \mathrm{l}) \times 10^{-3}}{600 \times 10^{-9}} \\
& =\frac{0.5 \times 0.032 \times 10^6}{600}=27 \text { fringes }
\end{aligned}
$$
Number of fringes shift due to extra path difference caused by transparent slab is
$$
n=\frac{(\mu-1) t}{\lambda}
$$
So, when two plates of thickness $t_1$ and $t_2$ are placed infront of slits, then shift is
$$
\begin{aligned}
n & =n_2-n_1=\frac{(\mu-1)\left(t_2-t_1\right)}{\lambda} \\
& =\frac{(1.5-1)(0132-0 \mathrm{l}) \times 10^{-3}}{600 \times 10^{-9}} \\
& =\frac{0.5 \times 0.032 \times 10^6}{600}=27 \text { fringes }
\end{aligned}
$$
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