In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, wh

Q: Why do we subtract the optical paths of the two plates?

Because they are in front of different slits. The upper plate increases the optical path of the ray from $S_1$, and the lower plate increases the optical path of the ray from $S_2$. The net path difference at the center is the difference between these two increases.

Q: How did we know the path difference is between $5\lambda$ and $5.5\lambda$?

The central maximum (where path difference is zero) has shifted downwards because the lower plate has a higher refractive index. The point $P$ (original center) now lies between the 5th maximum (path difference $5\lambda$) and the 6th minimum (path difference $5.5\lambda$). Since the 5th maximum is below $P$ and the 6th minimum is above $P$, the path difference at $P$ must be exactly between these two values.

Q: Why does the intensity formula use $\cos^2(\phi/2)$?

This comes from the superposition of two waves of equal amplitude. The resultant intensity $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\phi$. If $I_1 = I_2 = I_0$, this simplifies to $I = 2I_0(1 + \cos\phi) = 4I_0 \cos^2(\phi/2)$, where $4I_0$ is the maximum intensity $I_{\text{max}}$.

Q: What happens to the negative sign when solving $\cos(5\pi + \theta)$?

$\cos(5\pi + \theta)$ is indeed $-\cos\theta$. However, in the intensity formula, the cosine term is squared. Therefore, $(-\cos\theta)^2$ simply becomes $\cos^2\theta$, making the negative sign irrelevant for the intensity calculation.

Q: Can we neglect the absorption of light by the glass plates?

Yes, the problem explicitly states to neglect absorption. If absorption were considered, the amplitudes of the two interfering waves would no longer be equal, and the minimum intensity would not be zero, complicating the intensity formula.

Question

In a Young's experiment, the upper slit is covered by a thin glass plate of refractive index 1.4, while the lower slit is covered by another glass plate, having the same thickness as the first one but having refractive index 1.7. Interference pattern is observed using light of wavelength $5400 	ext{ \AA}$. It is found that the point $P$ on the screen, where the central maximum ($n = 0$) fall before the glass plates were inserted, now has $3/4$ the original intensity. It is further observed that what used to be the fifth maximum earlier lies below the point $P$ while the sixth minima lies above $P$. Calculate the thickness of glass plate. (Absorption of light by glass plate may be neglected).

Sigma Percentile · Accepted Answer

This is a classic and beautiful problem from Young's Double-Slit Experiment (YDSE) that tests your deep understanding of optical path difference, fringe shifting, and the relationship between phase difference and intensity. Let's embark on this journey to unravel the mystery of the shifting fringes!

### Analyzing the Setup

Imagine the standard YDSE setup. Light from two coherent sources, $S_1$ and $S_2$, interferes on a screen. Normally, the central maximum forms exactly at the geometric center, point $P$, because the path difference there is zero.

However, in our problem, we introduce a twist! We place a glass plate of refractive index $\mu_1 = 1.4$ in front of $S_1$ and another plate of refractive index $\mu_2 = 1.7$ in front of $S_2$. Both plates have the exact same thickness, $t$.

Because light travels slower in a denser medium, the optical path length increases. The ray from $S_2$ travels through a denser medium ($\mu_2 > \mu_1$) than the ray from $S_1$. Therefore, the optical path from $S_2$ to $P$ is longer than the optical path from $S_1$ to $P$.

The net path difference $\Delta x$ introduced at the central point $P$ is simply the difference in these optical paths:

$$\Delta x = (\mu_2 - \mu_1)t$$

Substituting the given values:

$$\Delta x = (1.7 - 1.4)t = 0.3 t$$

### Decoding the Fringe Shift

Because of this extra path difference, the entire fringe pattern shifts downwards (towards the slit with the higher refractive index plate). The problem gives us a fantastic clue: the original 5th maximum now lies *below* point $P$, and the 6th minimum lies *above* point $P$.

What does this mean mathematically?

At the 5th maximum, the path difference is exactly $5\lambda$. At the 6th minimum, the path difference is exactly $5\lambda + \frac{\lambda}{2} = 5.5\lambda$. Since point $P$ is sandwiched between these two, the path difference at $P$ must be strictly between $5\lambda$ and $5.5\lambda$.

We can elegantly express this as:

$$\Delta x = 5\lambda + \Delta$$

where $\Delta$ is some extra path length such that $\Delta < \frac{\lambda}{2}$.

### The Master Equation: Intensity and Phase

We are given that the intensity at point $P$ is now $\frac{3}{4}$ of the maximum intensity ($I_{	ext{max}}$). To use this, we need to convert our path difference into a phase difference, $\phi$. The fundamental relationship is:

$$\phi = \frac{2\pi}{\lambda} \Delta x$$

Substituting our expression for $\Delta x$:

$$\phi = \frac{2\pi}{\lambda} (5\lambda + \Delta) = 10\pi + \frac{2\pi}{\lambda}\Delta$$

Now, we bring in the master equation for interference intensity:

$$I = I_{	ext{max}} \cos^2 \left(\frac{\phi}{2}ight)$$

Plugging in our knowns:

$$\frac{3}{4} I_{	ext{max}} = I_{	ext{max}} \cos^2 \left(\frac{10\pi + \frac{2\pi}{\lambda}\Delta}{2}ight)$$

$$\frac{3}{4} = \cos^2 \left(5\pi + \frac{\pi}{\lambda}\Deltaight)$$

### Final Calculation

Let's simplify the trigonometric term. We know that $\cos(5\pi + 	heta) = -\cos	heta$. However, because the term is squared, the negative sign disappears:

$$\cos^2 \left(\frac{\pi}{\lambda}\Deltaight) = \frac{3}{4}$$

Taking the square root of both sides:

$$\cos \left(\frac{\pi}{\lambda}\Deltaight) = \frac{\sqrt{3}}{2}$$

This implies that the angle must be $\frac{\pi}{6}$:

$$\frac{\pi}{\lambda}\Delta = \frac{\pi}{6} \implies \Delta = \frac{\lambda}{6}$$

Notice how beautifully this fits our earlier condition! $\frac{\lambda}{6}$ is indeed less than $\frac{\lambda}{2}$.

Now, we can find the total path difference at $P$:

$$\Delta x = 5\lambda + \frac{\lambda}{6} = \frac{31\lambda}{6}$$

Finally, we equate this to our very first equation involving the thickness $t$:

$$0.3 t = \frac{31\lambda}{6}$$

$$t = \frac{31\lambda}{6 	imes 0.3}$$

Given the wavelength $\lambda = 5400 	ext{ \AA} = 5400 	imes 10^{-10} 	ext{ m}$:

$$t = \frac{31 	imes 5400 	imes 10^{-10}}{1.8}$$

$$t = 9.3 	imes 10^{-6} 	ext{ m}$$

**$$t = 9.3 \mu	ext{m}$$**

And there we have it! By carefully tracking the optical path and understanding the geometry of the shifted fringes, we've successfully deduced the thickness of the glass plates.

Visualized Solution

Path Difference at Central Point

Fringe Shift Analysis

Phase Difference at $P$

Intensity at $P$

Solving for $Δ$

Calculating Thickness $t$

The Way Forward

The Sigma Insight: Interference and Young's Double-Slit Experiment

Analyzing the Setup

Decoding the Fringe Shift

The Master Equation

Final Calculation

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