Question

In a YDSE two thin transparent sheets are used in front of the slits $S_1$ and $S_2$ having refractive indices $\mu_1=1.6$ and $\mu_2=1.4$ respectively.

If both sheets have thickness 't', the central maximum is observed at a distance of 5 mm from centre O. Now the sheets are replaced by two sheets of same material of refractive index $\frac{\mu_1+\mu_2}{2}$ but having thickness $t_1$ and $t_2$ such that $t'=\frac{t_1+t_2}{2}$. After replacement, now central maximum is observed at distance of 8 mm from centre O on the same side as before. Find the thickness $t_1$ (in $\mu m$).

[Given d = 1 mm, D = 1 m] (mark the answer in $\mu m$ to nearest integer)

Answer 1

33

Answer 2

The shift in the central maximum in a YDSE due to the insertion of thin transparent sheets of thicknesses $t_1$ and $t_2$ and refractive indices $\mu_1$ and $\mu_2$ in front of slits $S_1$ and $S_2$ respectively is given by

$y_c = \frac{D}{d} [(\mu_1 - 1)t_1 - (\mu_2 - 1)t_2]$.

Here, $y_c$ is the distance of the central maximum from the original center O. We assume the positive direction of $y$ is upwards. If $S_1$ is the upper slit and $S_2$ is the lower slit, a positive $y_c$ means the central maximum shifts upwards.

In the first scenario, sheets of thickness $t$ and refractive indices $\mu_1 = 1.6$ and $\mu_2 = 1.4$ are placed in front of $S_1$ and $S_2$ respectively. The central maximum is observed at $y_c = 5$ mm from O.

$5 \times 10^{-3} \text{ m} = \frac{1 \text{ m}}{1 \times 10^{-3} \text{ m}} [(1.6 - 1)t - (1.4 - 1)t]$

$5 \times 10^{-3} = 10^3 [0.6t - 0.4t]$

$5 \times 10^{-3} = 10^3 (0.2t)$

$5 \times 10^{-3} = 200t$

$t = \frac{5 \times 10^{-3}}{200} = \frac{5 \times 10^{-3}}{2 \times 10^2} = 2.5 \times 10^{-5} \text{ m} = 25 \times 10^{-6} \text{ m} = 25 \, \mu \text{m}$.

In the second scenario, the sheets are replaced by two sheets of the same material with refractive index $\mu = \frac{\mu_1 + \mu_2}{2} = \frac{1.6 + 1.4}{2} = 1.5$. The thicknesses are $t_1$ and $t_2$ in front of $S_1$ and $S_2$ respectively. The central maximum is observed at $y_c' = 8$ mm from O on the same side as before, so $y_c' = 8$ mm.

We are given that $t' = \frac{t_1 + t_2}{2} = t$. So, $t_1 + t_2 = 2t = 2 \times 25 \, \mu \text{m} = 50 \, \mu \text{m}$.

The shift of the central maximum is:

$y_c' = \frac{D}{d} [(\mu - 1)t_1 - (\mu - 1)t_2] = \frac{D}{d} (\mu - 1)(t_1 - t_2)$

$8 \times 10^{-3} \text{ m} = \frac{1 \text{ m}}{1 \times 10^{-3} \text{ m}} (1.5 - 1)(t_1 - t_2)$

$8 \times 10^{-3} = 10^3 (0.5)(t_1 - t_2)$

$8 \times 10^{-3} = 500(t_1 - t_2)$

$t_1 - t_2 = \frac{8 \times 10^{-3}}{500} = \frac{8 \times 10^{-3}}{5 \times 10^2} = 1.6 \times 10^{-5} \text{ m} = 16 \times 10^{-6} \text{ m} = 16 \, \mu \text{m}$.

We have a system of two linear equations for $t_1$ and $t_2$:

1. $t_1 + t_2 = 50 \, \mu \text{m}$

2. $t_1 - t_2 = 16 \, \mu \text{m}$

Adding the two equations:

$(t_1 + t_2) + (t_1 - t_2) = 50 + 16$

$2t_1 = 66$

$t_1 = \frac{66}{2} = 33 \, \mu \text{m}$.

Substituting $t_1 = 33$ into the first equation:

$33 + t_2 = 50$

$t_2 = 50 - 33 = 17 \, \mu \text{m}$.

The thickness $t_1$ is $33 \, \mu \text{m}$. We need to mark the answer in $\mu m$ to the nearest integer. Since 33 is an integer, the answer is 33.