A beaker contains water upto a height h1 and kerosene of height h2 above water so that the total hei

Q: Why do we add the apparent shifts instead of calculating a combined refractive index?

While you can calculate an effective refractive index for the combination, adding the individual shifts is mathematically simpler and physically intuitive because each layer independently shifts the image formed by the layer below it.

Q: Does the order of the liquids (water on top or kerosene on top) change the total apparent shift?

No, the total apparent shift depends only on the heights and refractive indices of the individual layers, not on their sequence. The addition of shifts is commutative.

Q: What is the difference between apparent depth and apparent shift?

Apparent depth is the distance from the surface to the virtual image ($h/\mu$), while apparent shift is the distance the image has moved from the actual object ($h - h/\mu$).

Q: What if we view the beaker from an angle instead of directly from above?

The formula $\Delta h = h(1 - 1/\mu)$ is an approximation valid only for near-normal viewing (paraxial rays). If viewed from a large angle, the apparent depth changes and depends on the viewing angle.

Q: Can the apparent shift ever be negative?

A negative shift would mean the object appears deeper than it is. This happens if you are in a denser medium looking into a rarer medium (like a fish looking at a bird), but not when looking from air into a liquid.

Question

A beaker contains water upto a height $h_1$ and kerosene of height $h_2$ above water so that the total height of (water + kerosene) is $(h_1 + h_2)$. Refractive index of water is $\mu_1$ and that of kerosene is $\mu_2$. The apparent shift in the position of the bottom of the beaker when viewed from above is

Options:

(A) $\left(1 - \frac{1}{\mu_1}\right)h_2 + \left(1 - \frac{1}{\mu_2}\right)h_1$

(B) $\left(1 + \frac{1}{\mu_1}\right)h_1 + \left(1 + \frac{1}{\mu_2}\right)h_2$

(C) $\left(1 - \frac{1}{\mu_1}\right)h_1 + \left(1 - \frac{1}{\mu_2}\right)h_2$

(D) $\left(1 + \frac{1}{\mu_1}\right)h_2 - \left(1 + \frac{1}{\mu_2}\right)h_1$

Sigma Percentile · Accepted Answer

### The Illusion of Depth: Apparent Shift in Multiple Liquids

Have you ever looked into a swimming pool and thought it was shallower than it actually is? This optical illusion is a classic consequence of refraction. When light travels from a denser medium (like water) to a rarer medium (like air), it bends away from the normal, causing the bottom to appear raised.

But what happens when you have multiple layers of different liquids, like oil floating on water? Does the math get incredibly complicated? Let's dive into the physics of apparent shift and see how beautifully simple it actually is!

### The Single Layer Shift

Before tackling multiple layers, let's recall the behavior of a single liquid layer. If you have a liquid of height $h$ and refractive index $\mu$, the apparent depth $h_{app}$ is given by $\frac{h}{\mu}$.

The **apparent shift** ($\Delta h$) is the difference between the actual depth and the apparent depth. Mathematically, it is expressed as:

$$\Delta h = h - h_{app} = h - \frac{h}{\mu} = h\left(1 - \frac{1}{\mu}\right)$$

This formula tells us exactly how much the bottom appears to be lifted upwards due to the presence of that specific liquid layer.

### Stacking the Layers

Now, imagine stacking a layer of kerosene on top of the water. The beauty of paraxial optics is that the apparent shifts produced by multiple immiscible layers are strictly additive. Each layer independently shifts the image formed by the layer below it.

For our beaker, we have two layers:
1. **Water Layer:** Height = $h_1$, Refractive Index = $\mu_1$
2. **Kerosene Layer:** Height = $h_2$, Refractive Index = $\mu_2$

The shift produced by the water layer alone is:

$$\Delta h_1 = h_1\left(1 - \frac{1}{\mu_1}\right)$$

Similarly, the shift produced by the kerosene layer is:

$$\Delta h_2 = h_2\left(1 - \frac{1}{\mu_2}\right)$$

### The Final Calculation

Since the layers are stacked one above the other, the total apparent shift of the bottom of the beaker is simply the sum of these two individual shifts:

$$\Delta h_{total} = \Delta h_1 + \Delta h_2$$

Substituting our expressions for $\Delta h_1$ and $\Delta h_2$, we get the final result:

**$$\Delta h_{total} = h_1\left(1 - \frac{1}{\mu_1}\right) + h_2\left(1 - \frac{1}{\mu_2}\right)$$**

This perfectly matches option (c). This principle is incredibly powerful because it generalizes to any number of layers. If you had $n$ immiscible liquids, you would just sum up $n$ individual shift terms!

Select Answer:

Visualized Solution

System Setup

Apparent Shift Formula

Shift due to Water

Shift due to Kerosene

Total Apparent Shift

Final Expression

Generalization

The Sigma Insight: Refraction and Total Internal Reflection

The Illusion of Depth

The Single Layer Shift

Stacking the Layers

The Final Calculation

Similar Questions

A vessel of depth $2 h$ is half filled with a liquid of refractive index $22$ and the upper half with another liquid of refractive index $2$ . The liquids are immiscible. The apparent depth of the inner surface of the bottom of vessel will be