In a Young's double slit experiment, a combination of two glass wedges A and B, having refractive in

Q: Why do we use similar triangles here?

Because the interface between the wedges is a straight diagonal line. This linear geometry means the thickness of each wedge varies linearly with its vertical position, making similar triangles the perfect tool to find the thickness at any specific height.

Q: Why is the total height of the wedge $4\text{ mm}$?

The problem states the distance between the slits is $2\text{ mm}$, and the distance from the slits to the top and bottom edges of the wedge block is $1\text{ mm}$ on both sides. Adding these up gives a total height of $1 + 2 + 1 = 4\text{ mm}$.

Q: Does the shift depend on the wavelength of light?

No, the position of the central maximum is independent of the wavelength. The central maximum is simply the point of zero total path difference, which is a purely geometric and refractive property of the setup.

Q: What if the wedges had the same refractive index?

If $\mu_A = \mu_B$, the entire block would act as a single glass slab of uniform thickness $t = 12\ \mu\text{m}$. It would introduce the exact same optical path difference for both slits, resulting in a net path difference of zero and no shift of the central maximum.

Q: Why do we subtract the path differences of the two slits?

We subtract them to find the net relative phase difference between the two interfering waves right after they exit the wedges. This net difference is what the geometric path difference must compensate for to form the central maximum.

Question

In a Young's double slit experiment, a combination of two glass wedges $A$ and $B$, having refractive indices $1.7$ and $1.5$, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $d = 2	ext{ mm}$ and the shortest distance between the slits and the screen is $D = 2	ext{ m}$. Thickness of the combination of the wedges is $t = 12\ \mu	ext{m}$. The value of $l$ as shown in the figure is $1	ext{ mm}$. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to O by ____

Sigma Percentile · Accepted Answer

### The Shifting Sands of Interference: YDSE with Glass Wedges

Welcome to a brilliant variation of the classic Young's Double Slit Experiment (YDSE). In the standard setup, light travels through the air, creating a beautiful, symmetric interference pattern on the screen. But what happens when we introduce a twist?

Here, we have placed a combination of two glass wedges, **A** and **B**, right in front of the slits. These wedges have different refractive indices, $\mu_A = 1.7$ and $\mu_B = 1.5$. This seemingly simple addition completely alters the optical landscape, forcing the central maximum to abandon its usual home at the center of the screen.

### The Optical Path: A Tale of Two Mediums

When light travels through a medium other than a vacuum (or air), it slows down. This means that a wave traveling through a glass wedge will cover a shorter physical distance in the same amount of time compared to a wave traveling through air.

To account for this, we use the concept of **optical path length**. If a light ray travels through a medium of thickness $t$ and refractive index $\mu$, the additional path difference introduced compared to traveling through air is given by the master equation:

$$\Delta x = (\mu - 1)t$$

In our setup, the light from each slit must pass through both wedge B and wedge A. The total extra optical path for a ray will be the sum of the extra paths introduced by each wedge.

### Geometry of the Wedges: Decoding the Thickness

To calculate the exact optical path, we need to know the precise thickness of wedge A and wedge B directly in front of each slit. This is where geometry comes to our rescue.

Let's look at the wedge combination. It forms a rectangular block of total width $t = 12\ \mu	ext{m}$. The diagonal line separates wedge B (on the left) from wedge A (on the right).

What is the total height of this block? The problem states that the distance between the slits is $d = 2	ext{ mm}$, and the distance from the slits to the top and bottom edges of the block is $l = 1	ext{ mm}$. Therefore, the total height is:

$$	ext{Total Height} = l + d + l = 1 + 2 + 1 = 4	ext{ mm}$$

Now, let's use similar triangles to find the thickness of wedge B (let's call it $x$) at the top slit $S_1$. The top slit is located $1	ext{ mm}$ from the top edge of the block.

$$\frac{x}{1	ext{ mm}} = \frac{12\ \mu	ext{m}}{4	ext{ mm}}$$

Solving this gives $x = 3\ \mu	ext{m}$. This means at $S_1$, the thickness of wedge B is $t_{B1} = 3\ \mu	ext{m}$. Since the total width is $12\ \mu	ext{m}$, the thickness of wedge A at $S_1$ is $t_{A1} = 12 - 3 = 9\ \mu	ext{m}$.

Similarly, for the bottom slit $S_2$, the distance from the top edge is $1	ext{ mm} + 2	ext{ mm} = 3	ext{ mm}$. Using similar triangles again, the thickness of wedge B at $S_2$ is $t_{B2} = 9\ \mu	ext{m}$, and the thickness of wedge A is $t_{A2} = 3\ \mu	ext{m}$.

### Calculating the Shift: Balancing the Paths

Now we can calculate the extra optical path for the rays emerging from both slits.

For the ray from $S_1$:
$$\Delta x_1 = (\mu_B - 1)t_{B1} + (\mu_A - 1)t_{A1}$$
$$\Delta x_1 = (1.5 - 1)(3) + (1.7 - 1)(9) = 1.5 + 6.3 = 7.8\ \mu	ext{m}$$

For the ray from $S_2$:
$$\Delta x_2 = (\mu_B - 1)t_{B2} + (\mu_A - 1)t_{A2}$$
$$\Delta x_2 = (1.5 - 1)(9) + (1.7 - 1)(3) = 4.5 + 2.1 = 6.6\ \mu	ext{m}$$

The net path difference created purely by the wedges is the difference between these two values:

$$\Delta x_{	ext{wedges}} = \Delta x_1 - \Delta x_2 = 7.8 - 6.6 = 1.2\ \mu	ext{m}$$

The central maximum is defined as the point on the screen where the **total path difference is zero**. Since the ray from $S_1$ has traveled a longer optical path through the wedges, it must travel a shorter geometric path to the screen to compensate.

This means the central maximum will shift upwards to a point $y$ where the geometric path difference $\frac{yd}{D}$ exactly balances the wedge path difference:

$$\frac{yd}{D} = \Delta x_{	ext{wedges}}$$

### The Final Result

We have all the pieces of the puzzle. Let's substitute the values and find the shift $y$:

$$y = \frac{D}{d} \Delta x_{	ext{wedges}}$$
$$y = \frac{2}{2 	imes 10^{-3}} 	imes 1.2 	imes 10^{-6}$$
$$y = 1.2 	imes 10^{-3}	ext{ m} = 1.2	ext{ mm}$$

And there we have it! The central maximum shifts by exactly **1.2 mm**. This problem beautifully demonstrates how geometric optics and wave optics intertwine to create fascinating physical phenomena.

Enter Numerical Value:

Visualized Solution

$Visual Anchor: YDSE with Wedges$

$Optical Path Difference$

$Total Height of Wedge$

$Thickness at Top Slit S_{1}$

$Thickness Distribution$

$Extra Optical Path$

$Calculating Path Differences$

$Net Path Difference$

$Condition for Central Maxima$

$Final Calculation$

$The Way Forward$

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

The Shifting Sands of Interference

The Optical Path

Geometry of the Wedges

Calculating the Shift

The Final Result

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.

Consider a Young's double slit experiment as shown in figure. What should be the slit separation $d$ in terms of wavelength $λ$ such that the first minima occurs directly in front of the slit ( $S_{1}$ )?

In a Young's double slit experiment, two slits are separated by $2 mm$ and the screen is placed one metre away. When a light of wavelength $500 nm$ is used, the fringe separation will be

Enter Numerical Value:

Visualized Solution

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

.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; } Total Height of Wedge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Thickness at Top Slit S1​

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

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

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

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

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

.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