In Young's double slit experiment using monochromatic light the fringe pattern shifts by a certain d

Q: Why does the fringe pattern shift when a slab is introduced?

The slab introduces an additional optical path length $(\mu - 1)t$ for the wave passing through it. To compensate for this and maintain a zero phase difference for the central maximum, the pattern shifts towards the slit covered by the slab.

Q: Does the fringe width change when the slab is introduced?

No, the fringe width $\omega = \frac{\lambda D}{d}$ depends only on the wavelength, screen distance, and slit separation. The slab only shifts the entire pattern without altering the distance between successive fringes.

Q: Why did we equate the shift to the new fringe width?

The problem explicitly states that the distance between successive maxima (which is the definition of fringe width) in the new setup (with screen at $2D$) is equal to the initial fringe shift caused by the mica sheet.

Q: What happens if the slab is placed in front of the other slit?

The magnitude of the shift remains exactly the same, but the direction of the shift reverses. The pattern always shifts towards the slit that is covered by the denser medium.

Q: Why did the D and d terms cancel out?

Both the fringe shift and the fringe width are proportional to $\frac{D}{d}$. When we equated the two expressions, this common geometric factor canceled out, leaving a relationship purely between the optical properties of the setup.

Question

In Young's double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index $1.6$ and thickness $1.964$ microns is introduced in the path of one of the interfering waves. The mica sheet is then removed and the distance between the slits and the screen is doubled. It is found that the distance between successive maxima (or minima) now is the same as the observed fringe shift upon the introduction of the mica sheet. Calculate the wavelength of the monochromatic light used in the experiment (in $	ext{\AA}$).

Sigma Percentile · Accepted Answer

The beauty of Young's Double Slit Experiment (YDSE) lies in its delicate sensitivity to phase differences. Even the slightest change in the optical path of one of the interfering waves can dramatically alter the interference pattern observed on the screen. In this problem, we explore exactly such a scenario, where a thin mica sheet acts as the disruptor, and we use the resulting shift to deduce the wavelength of the light itself.

### Analyzing the Setup

Imagine the standard YDSE setup. Two coherent light waves emerge from slits $S_1$ and $S_2$ and travel towards a screen at a distance $D$. When a transparent mica sheet of thickness $t$ and refractive index $\mu$ is introduced in front of one of the slits, it forces the light wave to travel through a denser medium.

Because light travels slower in the mica sheet compared to air, it takes longer for that specific wave to reach the screen. This introduces an additional optical path length of $(\mu - 1)t$. To compensate for this "delay" and maintain a zero phase difference for the central maximum, the entire fringe pattern shifts towards the slit covered by the sheet.

### The Master Equation

The physical distance by which the fringe pattern shifts on the screen is given by the well-known formula:

$$ \Delta y = \frac{(\mu - 1)tD}{d} $$

Here, $d$ is the separation between the two slits. Notice that this shift is independent of the wavelength of the light! It only depends on the properties of the slab and the geometry of the setup.

Next, the problem introduces a twist. The mica sheet is removed, restoring the original symmetry, but the screen is moved twice as far away, to a new distance of $2D$.

In this new configuration, the distance between successive maxima—which is simply the fringe width $\omega'$—is given by:

$$ \omega' = \frac{\lambda(2D)}{d} $$

### Equating and Solving

The core of the problem lies in a single, elegant condition: the initial fringe shift $\Delta y$ is exactly equal to the new fringe width $\omega'$. Let's set them equal to each other:

$$ \frac{(\mu - 1)tD}{d} = \frac{\lambda(2D)}{d} $$

Look at how beautifully the geometric factors $D$ and $d$ cancel out from both sides! This leaves us with a pure relationship between the optical properties:

$$ (\mu - 1)t = 2\lambda $$

Rearranging to solve for the unknown wavelength $\lambda$:

$$ \lambda = \frac{(\mu - 1)t}{2} $$

### Final Calculation

Now, it's just a matter of plugging in the given values. We know the refractive index $\mu = 1.6$ and the thickness $t = 1.964 	imes 10^{-6} 	ext{ m}$.

$$ \lambda = \frac{(1.6 - 1) 	imes 1.964 	imes 10^{-6}}{2} $$

$$ \lambda = \frac{0.6 	imes 1.964 	imes 10^{-6}}{2} $$

$$ \lambda = 0.3 	imes 1.964 	imes 10^{-6} $$

$$ \lambda = 0.5892 	imes 10^{-6} 	ext{ m} $$

To express this in a more standard unit for wavelength, we convert meters to Angstroms ($	ext{\AA}$). Since $1 	ext{ m} = 10^{10} 	ext{ \AA}$:

$$ \lambda = 0.5892 	imes 10^{-6} 	imes 10^{10} 	ext{ \AA} $$

$$ \lambda = 5892 	ext{ \AA} $$

And there we have it! By simply observing the macroscopic shift of fringes on a screen, we have successfully measured the microscopic wavelength of the light.

Enter Numerical Value:

Visualized Solution

$YDSE with a Slab$

$Fringe Shift Formula$

$Modified Setup$

$New Fringe Width$

$Equating Shift and Width$

$Solving for Wavelength$

$Substituting Values$

$Final Calculation$

$The Way Forward$

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

Analyzing the Setup

The Master Equation

Equating and Solving

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } YDSE with a Slab

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

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

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

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

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

.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; } Final Calculation

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