A thin paper of thickness 0.02mm having a refractive index 1.45 i… - Vidyadip

Question

A thin paper of thickness $0.02mm$ having a refractive index $1.45$ is pasted across one of the slits in a Young's double slit experiment. The paper transmits $\frac{4}{9}$ of the light energy falling on it.

Find the ratio of the maximum intensity to the minimum intensity in the fringe pattern.
How many fringes will cross through the centre if an identical paper piece is pasted on the other slit also? The wavelength of the light used is 600nm.

✓

Answer

Given that, $t = 0.02mm = 0.02 \times 10^{-3}m$, $\mu_1=1.45,\ \lambda=600\text{nm}=600\times10^{-9}\text{m}$

Let, $I_1$ = Intensity of source without paper = I
Then $I_2$ = Intensity of source with paper $=\Big(\frac{4}{9}\Big)\text{I}$

$\Rightarrow\frac{\text{I}_1}{\text{I}_2}=\frac{9}{4}\Rightarrow\frac{\text{r}_1}{\text{r}_2}=\frac{3}{2}\ [\because\text{I}\propto\text{r}^2]$
where, $r_1$ and $r_2$ are corresponding amplitudes.
So, $\frac{\text{I}_\text{max}}{\text{I}_\text{min}}=\frac{(\text{r}_1+\text{r}_2)^2}{(\text{r}_1-\text{r}_2)^2}=25:1$
No. of fringes that will cross the origin is given by,
$\text{n}=\frac{(\mu-1)\text{t}}{\lambda}=\frac{(1.45-1)\times0.02\times10^{-3}}{600\times10^{-9}}=15.$

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