YDSE with thin slab

Young's Double Slit Experiment (YDSE) is a classic demonstration of the wave nature of light, showcasing how light waves interfere to produce a pattern of bright and dark fringes. When a thin slab of material with a different refractive index is introduced between the slits and the screen, it alters the path difference between the interfering waves, thus shifting the interference pattern. This modification provides deeper insights into the principles of wave optics and the effect of mediums on light propagation. A real-life analogy can be seen in the way glasses correct our vision by altering the path of light entering our eyes, much like the slab adjusts the interference fringes. In this article, we will discuss the concept of Ydse With Thin Slab and provide examples for concept clarity.

YDSE with a Thin Slab

Young's Double-Slit Experiment is a classic demonstration of the wave nature of light, where an interference pattern of bright and dark fringes is produced on a screen by light passing through two closely spaced slits. When a thin transparent slab is introduced into the path of light from one of the slits, it alters the optical path length for the light passing through that slit. This results in a shift in the interference pattern.

Consider the arrangement of Young's double slit experiment as shown in Fig. In which a thin transparent film of refractive index $\mu$ and thickness 't' is introduced in front of the lower slit 'S'. Our aim is to obtain the new position of the nth maxima and minima. Let us assume a point P on screen at a distance Y from the origin O. It is important to note that in this particular situation, we cannot calculate the phase difference between the two waves arriving at directly by calculating the path difference $\left(S_2 P-S_1 P\right)$ because the two waves are not travelling in the same medium. The lower wave travels some distance in a medium $\mu$ and the remaining distance in air, while the upper wave travels all the distance in the air and travelled in the effective path difference we need to convert the distance travelled in medium $\mu$ into its equivalent distance in air, which is equal to $\mu$ and it is called the optical path. Hence the optical path is the equivalent distance to be travelled in the air to produce the same phase change as that produced in actual in traveling the actual distance. Thus, the optical path difference between the two waves is

$
\begin{aligned}
& \Delta \mathrm{x}=\left[\left(\mathrm{S}_2 \mathrm{P}-\mathrm{t}\right)+\mu \mathrm{t}\right]-\mathrm{S}_1 \mathrm{P}_{\text {or }} \\
& \Delta \mathrm{x}=\left(\mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}\right)+(\mu-1) \mathrm{t} \\
& \text { since } \quad \mathrm{S}_2 \mathrm{P}-\mathrm{S}_1 \mathrm{P}=\mathrm{d} \sin \theta=\mathrm{d}\left(\mathrm{y}^{\prime} / \mathrm{D}\right) \quad \text { (from the fig.) } \\
& \therefore \quad \Delta \mathrm{x}=\mathrm{dy}_{\mathrm{n}}^{\prime} / \mathrm{D}+(\mu-1) \mathrm{t}
\end{aligned}
$

From the nth maxima,
$
\begin{aligned}
& \Delta \mathrm{x}=\mathrm{n} \lambda, \therefore \mathrm{n} \lambda=\mathrm{dy}_{\mathrm{n}} / \mathrm{D}+(\mu-1) \mathrm{t} \text { or } \\
& y_n=\frac{n \lambda D}{d}-\frac{(\mu-1) t D}{d}
\end{aligned}
$

The position of the nth maxima and minima has shifted downward by the same distance which is called
$
\mathrm{S}=\mathrm{y}_{\mathrm{n}}-\mathrm{y}_{\mathrm{n}}^{\prime}=(\mu-1) \frac{\mathrm{tD}}{\mathrm{d}}
$

The distance between two successive maxima or minima remains unchanged. That is, the fringe width remains unchanged by introducing a transparent film.

The distance of shift is in the direction where the film is introduced. That is, if a film is placed in front of the upper slit the $S_1$ fringe pattern shifts upwards, if a film is placed in front of the lower slit $S_2$ the fringe pattern shifts downward.

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Solved Examples Based On YDSE with a Thin slab

Example 1: The figure shows Young's double-slit experimental setup. It is observed that when a thin transparent sheet of thickness $t$ and refractive index $\mu$ is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light is $\lambda, t$ will be :

1) $\frac{2 n D \lambda}{a(\mu-1)}$

2) $\frac{n \lambda}{\mu-1}$

3) $\frac{D \lambda}{a(\mu-1)}$

4) $\frac{2 D \lambda}{a(\mu-1)}$

Solution:

Displacement of fringe

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

Shift in the position of fringes

wherein

If the sheet is placed in front of one of the slits

If the sheet is placed in front of one slit,

$
\begin{aligned}
\text { displacement of firing } & =\frac{D}{d}(\mu-1) t \\
\text { given, } \quad & \frac{n \lambda D}{d}=\frac{D}{d}(\mu-1) t \\
& t=\frac{n \lambda}{\mu-1}
\end{aligned}
$
Hence, the answer is the option (2).

Example 2: A thin glass of thickness $\frac{2500}{3} \lambda$ (the wavelength of light used) and refractive index $\mu=1.5$ is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :

1) $2: 1$

2) $1: 4$

3) $4: 1$

4) $4: 3$

Solution:

Due to the introduction of the sheet in front of one slit (thickness t and refractive index $(\mu)$ the shift $=\frac{D}{d}(\mu-1) t$

i.e., the path difference becomes $(\mu-1) t$ instead of zero at the centre of the screen i.e at the centre $(\Delta x \neq 0)$

therefore Phase difference $=\frac{2 \pi}{\lambda} \times(\mu-1) t$

So using So using $t=\frac{2500}{3} \lambda_{\text {we get }} \Delta \phi=\frac{2 \pi}{\lambda} \times \frac{2500}{3} \lambda \times 0.5=\frac{1250}{3} \times 2 \pi$ at the centre.

The intensity at the centre will same as that of the point having the Phase difference as $\frac{2 \pi}{3}$

i.e 120⁰

So Let the new point where the Phase difference is 0 be A

So another point B which is equidistant from the slits as A will have a phase difference

So

where $I_o$ is the maximum intensity or Intensity of a point when the phase difference is 0.

So $I_A=I_0$

So $I_B=I_0 \operatorname{Cos}^2\left(\frac{4 \pi}{3}\right)=\frac{I_0}{4}$

So, $\frac{I_A}{I_B}=4$

Hence, the answer is the option (3).

Example 3: In Young's double slit experiment performed using monochromatic light of wavelength $\lambda$, when a glass plate $(\mu=1.5)$ of thickness $\mathrm{x} \lambda$ is introduced in the path of one of the interfering beams, the intensity at the position where the central maximum occurred previously remains unchanged. The value of x will be :

1)3

2)2

3)1.5

4)0.5

Solution:

For the intensity to remain the same the position must be of maxima.

'So path difference must be $n \lambda$

$\begin{aligned} & (1.5-1) x \lambda=n \lambda \\ & x=2 n, n=0,1,2 \ldots \\ & x=0,2,4,6 \ldots\end{aligned}$

Hence, the answer is the option (2).

Example 4: A thin mica sheet of thickness $4 \times 10^{-6} \mathrm{~m}$ and refractive index $(\mu=1.5)$ is introduced in the path of the light from the upper slit. The wavelength of the wave used in $5000 $ Angstrom. The central bright maximum will shift:

1)4 fringes upward

2)2 fringes downward

3)10 fringes upward

4)none of these

Solution:

On introducing the mica sheet, fringe shift $=\frac{\beta}{\lambda}(\mu-1) \mathrm{t}$

Where t is the thickness of the sheet.

So, shift $=\frac{\beta}{\left(5000 \times 10^{-10}\right)} \times(1.5-1) \times\left(4 \times 10^{-6}\right)=4 \beta$

So, the central bright maximum will shift 4 fringes upward.

Hence, the answer is the option (1).

Summary

In Young's Double Slit Experiment with a thin slab, a transparent material of a certain thickness and refractive index is placed in front of one of the slits. The slab creates a delay in the light, thus changing the interference pattern on the screen. The slab modifies the optical path length, which leads to a distribution of the light in the form of a bright and dark fringe. It is due to the fact that if half a wave is added by the slab, the waves are one identity away and then if when the bright band moves away from the screen, it shows the effect of the optical path difference on the formation of waves. We can see how the adjustment of the optical path length affects the pattern of a single fringe because the slab increases it by half a wavelength.