A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index o

Q: Why did we use $h = 7.2 \text{ m}$ instead of $12.8 \text{ m}$ in the kinematics equation?

The equation $v^2 = u^2 + 2gh$ uses the distance the object has *fallen* from its starting point, not its height above the ground. Since it started at $20 \text{ m}$ and is now at $12.8 \text{ m}$, it has fallen $20 - 12.8 = 7.2 \text{ m}$.

Q: Why is the apparent height $\mu x$ and not $\frac{x}{\mu}$?

When looking from a denser medium (water) into a rarer medium (air), the refracted rays bend towards the normal. When extended backwards, they meet at a point higher than the actual object. Hence, the apparent height is greater, given by $\mu x$.

Q: Does the apparent velocity depend on the actual height $x$?

No! As we saw in the derivation, $v_{\text{app}} = \mu v$. The apparent velocity only depends on the refractive index and the actual velocity at that instant, regardless of the height $x$.

Q: What if the fish was in the air looking at a sinking stone in the water?

In that case, the observer is in the rarer medium looking into the denser medium. The apparent depth would be $\frac{x}{\mu}$, and the apparent velocity would be $\frac{v}{\mu}$.

Q: Why did we take $g = 10 \text{ ms}^{-2}$ instead of $9.8 \text{ ms}^{-2}$?

In JEE problems, unless specified otherwise, taking $g = 10 \text{ ms}^{-2}$ often leads to clean, integer calculations (like $\sqrt{144} = 12$). If the options were very close, we might need to use $9.8$, but here $10$ works perfectly.

Question

A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees the speed of ball as

Options:

(A) $9 	ext{ ms}^{-1}$

(B) $12 	ext{ ms}^{-1}$

(C) $16 	ext{ ms}^{-1}$

(D) $21.33 	ext{ ms}^{-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing by a serene lake. You drop a ball from a height of $20 	ext{ m}$ above the water surface. Deep inside the lake, a fish is looking straight up, watching the ball fall. We are asked to find the speed of the ball as it appears to the fish at the exact moment the ball is $12.8 	ext{ m}$ above the water surface.

This problem beautifully intertwines two fundamental concepts of physics: **Kinematics** (the motion of the ball) and **Ray Optics** (how light travels between different media).

### The Master Equation for Kinematics

First, let's figure out how fast the ball is actually moving. The ball was dropped, which means its initial velocity $u = 0$.

At the instant in question, the ball is $12.8 	ext{ m}$ above the water. Since it started from $20 	ext{ m}$, the distance it has fallen is:
$$h = 20 	ext{ m} - 12.8 	ext{ m} = 7.2 	ext{ m}$$

Using the third equation of motion, we can find its actual velocity $v$:
$$v^2 = u^2 + 2gh$$
$$v = \sqrt{2gh}$$

Substituting $g = 10 	ext{ ms}^{-2}$ and $h = 7.2 	ext{ m}$:
$$v = \sqrt{2 	imes 10 	imes 7.2} = \sqrt{144} = 12 	ext{ ms}^{-1}$$

So, the ball is actually falling at $12 	ext{ ms}^{-1}$.

### The Illusion of Optics

Now, let's step into the perspective of the fish. The fish is in water (a denser medium with refractive index $\mu = \frac{4}{3}$), looking at the ball in the air (a rarer medium).

When light from the ball enters the water, it bends towards the normal. Because of this refraction, the fish perceives the ball to be higher than it actually is. The apparent height $x_{	ext{app}}$ of an object in a rarer medium viewed from a denser medium is given by:
$$x_{	ext{app}} = \mu x$$

where $x$ is the actual height of the ball above the water surface.

### Final Calculation: Apparent Velocity

We don't just want the apparent position; we want the apparent speed. Speed is the rate of change of position. So, let's differentiate our apparent height equation with respect to time $t$:
$$\frac{dx_{	ext{app}}}{dt} = \mu \frac{dx}{dt}$$

Here, $\frac{dx_{	ext{app}}}{dt}$ is the apparent velocity $v_{	ext{app}}$, and $\frac{dx}{dt}$ is the actual velocity $v$.
$$v_{	ext{app}} = \mu v$$

Substitute the values we know:
$$v_{	ext{app}} = \frac{4}{3} 	imes 12 	ext{ ms}^{-1}$$
$$v_{	ext{app}} = 16 	ext{ ms}^{-1}$$

The fish sees the ball hurtling towards it at a terrifying $16 	ext{ ms}^{-1}$, even though it's only moving at $12 	ext{ ms}^{-1}$! This is a classic example of how refraction not only distorts position but also perceived motion.

Select Answer:

Visualized Solution

Visualizing the Setup

Actual Speed of the Ball

Calculating Actual Speed

Apparent Position

Apparent Velocity

Final Calculation

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Master Equation for Kinematics

The Illusion of Optics

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Setup

Actual Speed of the Ball

Calculating Actual Speed

Apparent Position

Apparent Velocity

Final Calculation

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Master Equation for Kinematics

The Illusion of Optics

Final Calculation

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