Question

A YDSE set-up is immersed in a liquid whose refractive index varies with time as $\mu=\frac{5}{2}-\frac{t}{4}$ until it reaches a steady state value of $\frac{5}{4}$. A glass plate of thickness 36 µm and refractive index $\frac{3}{2}$ is introduced in front of one of the slits. Find the time (in seconds) at which central maxima is at O.

Answer 1

4

Answer 2

The central maximum in a Young's Double Slit Experiment (YDSE) occurs at a point where the optical path difference between the waves from the two slits is zero.

1. Optical Path Difference (OPD) at point O: When a glass plate of thickness $T$ and refractive index $\mu_g$ is introduced in front of one of the slits (say $S_2$), and the entire setup is immersed in a liquid of refractive index $\mu$, the optical path difference at a point P on the screen is given by: $\Delta x = \mu (r_2 - r_1) + (\mu_g - \mu)T$ where $r_1$ and $r_2$ are the geometric path lengths from slits $S_1$ and $S_2$ to point P, respectively.

   For the central maximum to be at point O (the geometric center of the screen, directly opposite the midpoint of the slits), the geometric path lengths from $S_1$ and $S_2$ to O are equal, i.e., $r_1 = r_2$. Therefore, $r_2 - r_1 = 0$.

   Setting the optical path difference to zero for the central maximum at O: $\mu (0) + (\mu_g - \mu)T = 0$ $(\mu_g - \mu)T = 0$

2. Solve for the condition of refractive indices: Since the thickness of the glass plate $T = 36 \, \mu\text{m}$ is not zero, for the product to be zero, the term $(\mu_g - \mu)$ must be zero. $\mu_g - \mu = 0$ $\mu_g = \mu$

3. Substitute the given values and solve for time: We are given:

   • Refractive index of the liquid: $\mu = \frac{5}{2} - \frac{t}{4}$
   • Refractive index of the glass plate: $\mu_g = \frac{3}{2}$

   Equating $\mu_g$ and $\mu$: $\frac{3}{2} = \frac{5}{2} - \frac{t}{4}$

   Rearrange the equation to solve for $t$: $\frac{t}{4} = \frac{5}{2} - \frac{3}{2}$ $\frac{t}{4} = \frac{5 - 3}{2}$ $\frac{t}{4} = \frac{2}{2}$ $\frac{t}{4} = 1$ $t = 4 \, \text{seconds}$

4. Check validity: The problem states that the refractive index varies until it reaches a steady state value of $\frac{5}{4}$. Let's find the time when this occurs: $\frac{5}{4} = \frac{5}{2} - \frac{t}{4} \Rightarrow \frac{t}{4} = \frac{5}{2} - \frac{5}{4} = \frac{10 - 5}{4} = \frac{5}{4} \Rightarrow t = 5$ seconds. Since $t=4$ seconds is less than $5$ seconds, the formula for $\mu(t)$ is valid at this time.