A glass tumbler having inner depth of 17.5 cm is kept on a table. A student starts pouring water (mu

Q: Why do we use apparent depth for water but not for the empty part?

The empty part is filled with air, which has a refractive index of approximately 1. Therefore, light traveling through it doesn't bend significantly, and its apparent depth is equal to its actual depth.

Q: Does the width of the tumbler affect the answer?

No, the apparent depth formula $H_{app} = \frac{H}{\mu}$ is independent of the width or shape of the container, as long as we are looking from almost directly above (paraxial approximation).

Q: Why did the book's solution say $H = 10.11\text{ cm}$?

That was a calculation error in the reference material! $17.5 \times 4$ is exactly 70, and $70 / 7$ is exactly 10. Always trust your own math if you've set up the physics correctly.

Q: What if the observer looks from an angle instead of directly above?

If viewed from a significant angle, the apparent depth formula becomes more complex and depends on the viewing angle due to the geometry of refraction. The simple $H/\mu$ formula assumes near-normal viewing.

Q: Can the tumbler ever look completely full if it's not?

No, because the apparent depth is always a fraction of the actual depth (since $\mu > 1$). The water surface will always appear below the rim unless it's actually at the rim.

Question

A glass tumbler having inner depth of $17.5	ext{ cm}$ is kept on a table. A student starts pouring water ($\mu = 4/3$) 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?

Options:

(A) $11.7	ext{ cm}$

(B) $10	ext{ cm}$

(C) $7.5	ext{ cm}$

(D) $8.75	ext{ cm}$

Sigma Percentile · Accepted Answer

Have you ever looked into a swimming pool and thought the bottom was much closer than it actually is? Or noticed how a straw looks bent when placed in a glass of water? These everyday illusions are the handiwork of **refraction**—the bending of light as it passes from one medium to another.

In this fascinating problem, we encounter a student who is pouring water into a glass tumbler and relying purely on visual perception to judge when it is "half-filled." Let's dive into the physics of why our eyes deceive us and how we can use mathematics to uncover the truth.

### The Illusion of Depth

Imagine a glass tumbler with a total inner depth of $17.5	ext{ cm}$. We start pouring water into it up to an unknown actual height, which we will call $H$. Naturally, the remaining empty part of the tumbler, which is filled with air, has a height of $17.5 - H$.

When the student looks down into the tumbler from above, light rays traveling from the bottom of the water into the air bend away from the normal. Because our brains assume light travels in straight lines, we trace these bent rays backward, making the bottom of the tumbler appear higher than it truly is.

This perceived depth is known as the **apparent depth** ($H_{	ext{app}}$). The relationship between the actual depth and the apparent depth is governed by the refractive index ($\mu$) of the liquid:

$$H_{	ext{app}} = \frac{H}{\mu}$$

For water, the refractive index is given as $\mu = \frac{4}{3}$. Therefore, the apparent depth of the water is:

$$H_{	ext{app}} = \frac{H}{4/3} = \frac{3H}{4}$$

### Translating Perception into Math

The crux of the problem lies in a single phrase: *"When he feels that the tumbler is half filled..."*

What does it mean to "feel" or "see" that a container is half-filled from above? It means that the empty portion of the tumbler looks exactly equal in depth to the filled portion.

Since the empty part is filled with air ($\mu \approx 1$), its apparent depth is practically identical to its actual depth, which is $17.5 - H$. However, the water portion appears to have a depth of $\frac{3H}{4}$.

Equating these two perceived depths gives us our master equation:

$$17.5 - H = \frac{3H}{4}$$

### The Final Reveal

Now, we just need to solve this linear equation for $H$. Let's group the $H$ terms on one side by adding $H$ to both sides:

$$17.5 = H + \frac{3H}{4}$$

To add these, we find a common denominator:

$$17.5 = \frac{4H}{4} + \frac{3H}{4}$$

$$17.5 = \frac{7H}{4}$$

Finally, we isolate $H$ by multiplying by $4$ and dividing by $7$:

$$H = \frac{17.5 	imes 4}{7}$$

$$H = \frac{70}{7} = 10	ext{ cm}$$

And there we have it! Even though the student *perceives* the tumbler to be half-filled, the actual height of the water is $10	ext{ cm}$, which is more than half of the $17.5	ext{ cm}$ tumbler. Refraction has successfully tricked the observer, but it cannot trick the math!

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Concept of Apparent Depth$

$The ’Half-Filled’ Condition$

$Setting up the Equation$

$Simplifying the Equation$

$Solving for H$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Refraction and Total Internal Reflection

The Illusion of Depth

Translating Perception into Math

The Final Reveal

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

A large glass slab $(μ = \frac{5}{3})$ of thickness 8 cm is placed over a point source of light on a plane surface. It is seen that light emerges out of the top surface of the slab from a circular area of radius $R$ cm. What is the value of $R$ ?

Select Answer:

Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The ’Half-Filled’ Condition

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

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

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

.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