The Young's double slit experiment is done in a medium of refractive index 4/3. A light of 600 nm wa

Q: Why do we use relative refractive index $\frac{\mu_g}{\mu_m}$ instead of just $\mu_g$?

Because the glass slab is placed in a medium, not in a vacuum. The extra optical path is relative to the medium it replaces.

Q: Why does the central maximum shift downwards?

The glass slab is placed in front of the lower slit $S_2$. It slows down the light, increasing its optical path. To compensate and have zero net path difference, the geometrical path from $S_1$ must be longer, which happens below the central axis.

Q: How did we get the formula for geometrical path difference $\frac{yd}{D}$?

For small angles, the path difference is $d \sin\theta$. Since $\tan\theta = \frac{y}{D}$ and for small angles $\sin\theta \approx \tan\theta$, we get $\frac{yd}{D}$.

Q: Why is the phase difference formula $\frac{2\pi}{\lambda} \Delta x$?

A path difference of one full wavelength $\lambda$ corresponds to a phase difference of one full cycle, which is $2\pi$ radians. So, we multiply the path difference by $\frac{2\pi}{\lambda}$.

Q: Why do we only select wavelengths between 400 nm and 700 nm?

The problem asks for the wavelengths of white light that form maxima. White light consists of the visible spectrum, which typically ranges from 400 nm to 700 nm.

Question

The Young's double slit experiment is done in a medium of refractive index $4/3$. A light of $600	ext{ nm}$ wavelength is falling on the slits having $0.45	ext{ mm}$ separation. The lower slit $S_2$ is covered by a thin glass sheet of thickness $10.4	ext{ \mu m}$ and refractive index $1.5$. The interference pattern is observed on a screen placed $1.5	ext{ m}$ from the slits as shown in the figure.
(a) Find the location of central maximum (bright fringe with zero path difference) on the $y$-axis.
(b) Find the light intensity of point $O$ relative to the maximum fringe intensity.
(c) Now, if $600	ext{ nm}$ light is replaced by white light of range $400$ to $700	ext{ nm}$, find the wavelengths of the light that form maxima exactly at point $O$.
(All wavelengths in the problem are for the given medium of refractive index $4/3$. Ignore dispersion)

Sigma Percentile · Accepted Answer

The Young's Double Slit Experiment (YDSE) is a classic demonstration of the wave nature of light. But what happens when we introduce a medium and a glass slab into the mix? The problem transforms from a simple geometry exercise into a fascinating exploration of optical paths and phase differences. Let's dive into this thrilling puzzle!

### Analyzing the Setup

Imagine the standard YDSE setup, but instead of air, the entire apparatus is submerged in a medium with a refractive index of $4/3$. To make things even more interesting, a thin glass sheet of refractive index $1.5$ is placed right in front of the lower slit, $S_2$.

Because of this glass sheet, the light ray emerging from $S_2$ will have to travel an extra optical path. This path difference depends on the thickness of the slab and the relative refractive index of the glass with respect to the surrounding medium.

### The Downward Shift

Let's assume that the new central maximum is formed at a point $P$, which is a distance $y$ below the central point $O$. The geometrical path difference between the rays from $S_1$ and $S_2$ reaching this point is given by $\frac{yd}{D}$.

**There is a catch here.** For the central maximum, the net path difference must be exactly zero. This means the geometrical path difference and the optical path difference introduced by the glass sheet must perfectly cancel each other out.

Equating the two, we get:
$$ \frac{yd}{D} = \left(\frac{\mu_g}{\mu_m} - 1ight)t $$

By substituting the given values—the refractive indices, thickness, slit separation, and screen distance—we can solve for $y$:
$$ y = \left(\frac{1.5}{4/3} - 1ight) \frac{10.4 	imes 10^{-6} 	imes 1.5}{0.45 	imes 10^{-3}} $$

Upon calculating, the value of $y$ comes out to be $4.33 	imes 10^{-3} 	ext{ m}$, or $4.33 	ext{ mm}$. This means the central maximum shifts downwards by exactly this distance!

### Intensity at the Origin

Now let's move to the second part of our puzzle. We need to find the intensity at point $O$. At point $O$, the geometrical path difference is zero, so whatever path difference exists will be solely due to the glass sheet.

Let's convert this path difference into a phase difference. The phase difference $\phi$ is $\frac{2\pi}{\lambda}$ times the path difference:
$$ \phi = \frac{2\pi}{6 	imes 10^{-7}} \left(\frac{1.5}{4/3} - 1ight) (10.4 	imes 10^{-6}) $$

Solving this, the phase difference comes out to be $\frac{13\pi}{3}$.

Now recall the formula for intensity. The intensity is the maximum intensity times $\cos^2\left(\frac{\phi}{2}ight)$:
$$ I = I_{\max} \cos^2\left(\frac{13\pi}{6}ight) $$

The value of $\cos\left(\frac{13\pi}{6}ight)$ is $\frac{\sqrt{3}}{2}$, and its square will be $\frac{3}{4}$. So the intensity at point $O$ becomes **$\frac{3}{4} I_{\max}$**.

### The Colors of Interference

In the final part, we replace the monochromatic light with white light. We need to find the wavelengths that will form a maxima exactly at point $O$. For a maxima, the path difference must be an integral multiple of $\lambda$.

We have already found the expression for the path difference at point $O$. Let's calculate its exact value:
$$ \Delta x = \left(\frac{1.5}{4/3} - 1ight) (10.4 	imes 10^{-6} 	ext{ m}) = 1300 	ext{ nm} $$

Now we find the possible wavelengths by dividing this path difference by an integer $n$:
$$ \lambda = \frac{1300}{n} $$

For $n = 1$, it's $1300 	ext{ nm}$. For $n = 2$, it's $650 	ext{ nm}$. For $n = 3$, it's $433.33 	ext{ nm}$, and so on.

The visible range is given as $400 	ext{ nm}$ to $700 	ext{ nm}$. From our list, the wavelengths that fall perfectly into this range are **$650 	ext{ nm}$** and **$433.33 	ext{ nm}$**.

And there we have it! By carefully tracking the optical path and understanding the conditions for interference, we've successfully unraveled the mysteries of this complex setup.

Visualized Solution

$Setup and Given Data$

$Optical Path Difference$

$Geometrical Path Difference$

$Condition for Central Maximum$

$Substituting Values$

$Calculating Shift y$

$Path Difference at O$

$Phase Difference at O$

$Calculating Phase Difference$

$Intensity Formula$

$Calculating Intensity at O$

$Condition for Maxima with White Light$

$Exact Path Difference$

$Possible Wavelengths$

$Wavelengths in Visible Range$

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

Analyzing the Setup

The Downward Shift

Intensity at the Origin

The Colors of Interference

Similar Questions

In a Young's double slit experiment, two wavelengths of $500 nm$ and $700 nm$ were used. What is the minimum distance from the central maximum where their maximas coincide again? Take $D / d = 1 0^{3}$ . Symbols have their usual meanings.

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup and Given Data

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Optical Path Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Geometrical Path Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Central Maximum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Shift y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Path Difference at O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Phase Difference at O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Phase Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Intensity Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Intensity at O

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Maxima with White Light

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Exact Path Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Possible Wavelengths

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wavelengths in Visible Range