A vessel of depth 2h is half filled with a liquid of refractive index 2√2 and the upper half with an

Q: Why do we just add the apparent depths of each layer?

Because the shift produced by each layer is independent of the others. The total shift in the image position is the sum of the shifts produced by each individual layer.

Q: Does the order of the liquids matter?

No, the total apparent depth depends only on the thickness and refractive index of each layer, not on their sequence, as long as you are looking normally (perpendicularly) through them.

Q: What if we look at an angle instead of straight down?

If viewed at an angle, the apparent depth changes and depends on the angle of viewing. The formula $d/\mu$ is an approximation valid only for near-normal viewing (paraxial rays).

Q: Why did we multiply by $\sqrt{2}/\sqrt{2}$ at the end?

This is called rationalizing the denominator. It's a standard mathematical practice to remove square roots from the denominator, which helps in matching the final expression with the given options.

Q: What if the liquids were miscible?

If the liquids mix and form a continuous gradient of refractive index, we can't use the simple summation formula. We would need to integrate the differential apparent depth over the total depth.

Question

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

Options:

(A) $\frac{h}{\sqrt{2}}$

(B) $\frac{3}{4}h\sqrt{2}$

(C) $\frac{h}{3\sqrt{2}}$

(D) $\frac{h}{2(\sqrt{2}+1)}$

Sigma Percentile · Accepted Answer

### The Illusion of Depth: Apparent Depth in Multiple Liquid Layers

Imagine looking down into a vessel filled with two different immiscible liquids. The bottom half is filled with a denser liquid, and the top half with a lighter one. Because light bends as it travels from a denser medium to a rarer medium (like air), the bottom of the vessel appears to be raised. This phenomenon is known as apparent depth.

When dealing with a single liquid, the apparent depth is simply the real depth divided by the refractive index. But what happens when we have multiple layers?

### The Master Equation

When we look through multiple layers of different refractive indices, the total apparent depth is simply the sum of the apparent depths of each individual layer. The formula is elegantly simple:

$$ d_{	ext{app}} = \sum \frac{d_i}{\mu_i} $$

For our specific problem, we have two layers. Therefore, the equation becomes:

$$ d_{	ext{app}} = \frac{d_1}{\mu_1} + \frac{d_2}{\mu_2} $$

### Substituting the Values

Let's substitute our given values into the equation. The bottom layer has a depth of $h$ and a refractive index of $2\sqrt{2}$. The top layer also has a depth of $h$, but its refractive index is $\sqrt{2}$.

$$ d_{	ext{app}} = \frac{h}{2\sqrt{2}} + \frac{h}{\sqrt{2}} $$

To add these fractions, we need a common denominator, which is $2\sqrt{2}$. We multiply the numerator and denominator of the second term by $2$.

$$ d_{	ext{app}} = \frac{h + 2h}{2\sqrt{2}} $$

Adding the terms in the numerator, we get:

$$ d_{	ext{app}} = \frac{3h}{2\sqrt{2}} $$

### Final Calculation and Rationalization

Now, let's look at our options. They are in a slightly different form. It is a standard mathematical practice to rationalize the denominator by multiplying the numerator and denominator by $\sqrt{2}$.

$$ d_{	ext{app}} = \frac{3h}{2\sqrt{2}} 	imes \frac{\sqrt{2}}{\sqrt{2}} $$

This gives us our final, elegant result:

$$ d_{	ext{app}} = \frac{3\sqrt{2}h}{4} = \frac{3}{4}h\sqrt{2} $$

And there we have it! The apparent depth of the vessel's bottom is $\frac{3}{4}h\sqrt{2}$. This perfectly matches option (b).

**Food for Thought:** What if the liquids were miscible and formed a continuous gradient of refractive index? In that case, we couldn't use the simple summation formula. We would have to use integration to find the apparent depth. Think about how you would set up that integral!

Animated Solution for Physics - Optics: 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

Select Answer:

Visualized Solution

$A pp a r e n t D e pt h$

$F or m u l a f or A pp a r e n t D e pt h$

$S u b s t i t u t i o n$

$T ak in g L C M$

$S im pl i f i c a t i o n$

$R a t i o na l i z a t i o n$

$F ina l A n s w er$

$F oo df or T h o ug h t$

The Sigma Insight: Refraction and Total Internal Reflection

The Illusion of Depth

The Master Equation

Substituting the Values

Final Calculation and Rationalization

Similar Questions

A glass tumbler having inner depth of $17.5 cm$ is kept on a table. A student starts pouring water ( $μ = 4/3$ ) into it while looking at the surface of water from the above. When he feels that the tumbler is half filled, he stops pouring water. Up to what height, the tumbler is actually filled?

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Visualized Solution

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

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