An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the

Q: Why does the line of sight remain fixed?

Because the observer is looking through a pin-hole, which acts as a tiny aperture. To see anything, the light rays must pass exactly through this point and enter the eye at the specific angle dictated by the pin-hole's placement.

Q: How do we know the ray refracts exactly at the midpoint of the liquid surface?

The line of sight in the air is a straight line connecting the pin-hole $(2h, 3h)$ and the top of the rod $(0, h)$. The liquid surface is at height $2h$, which is exactly halfway vertically between $h$ and $3h$. By similar triangles, the horizontal position must also be exactly halfway, which is at distance $h$.

Q: Why do we take the angle with the vertical?

In optics, angles of incidence and refraction are always measured with respect to the normal to the surface. Since the liquid surface is horizontal, the normal is vertical.

Q: What if the beaker was filled to a different height?

If the liquid height changes, the point of intersection with the fixed line of sight would shift. This would change the required angle of the ray inside the liquid, meaning the observer might see a different part of the rod, or the refractive index would need to be different to see the bottom.

Q: Can we solve this using apparent depth?

Apparent depth formulas are typically derived for paraxial rays (rays very close to the normal). Here, the angles are large ($45^\circ$), so the standard apparent depth formula $d' = d/\mu$ is not accurate. We must use the exact geometry and Snell's Law.

Question

An observer can see through a pin-hole the top end of a thin rod of height $h$, placed as shown in the figure. The beaker height is $3h$ and its radius $h$. When the beaker is filled with a liquid up to a height $2h$, he can see the lower end of the rod. Then the refractive index of the liquid is

Options:

(A) $\frac{5}{2}$

(B) $\sqrt{\frac{5}{2}}$

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

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### The Power of a Fixed Line of Sight

Imagine you are peering through a tiny pin-hole into an empty beaker. The pin-hole acts as a strict gatekeeper—it only allows light rays traveling along one specific, unyielding straight line to enter your eye.

Initially, without any liquid, you can see the top of a thin rod placed inside the beaker. This tells us something crucial: the straight line connecting the top of the rod to the pin-hole is your **locked line of sight**. Any light that wishes to reach your eye, whether the beaker is empty or filled with liquid, must ultimately travel along this exact path once it enters the air.

### Analyzing the Initial Geometry

Let's break down the geometry of this locked line of sight. We can set up a coordinate system where the bottom-left corner of the beaker is the origin $(0,0)$.

The rod has a height of $h$, so its top is at the point $R(0, h)$. The pin-hole is located at the top-right corner of the beaker. Given the beaker's total height is $3h$ and its width is $2h$, the pin-hole is at the point $P(2h, 3h)$.

If we draw a right-angled triangle to find the slope of this line, the horizontal distance is $2h - 0 = 2h$, and the vertical distance is $3h - h = 2h$. Because the horizontal and vertical distances are perfectly equal, the line of sight makes an angle of exactly $45^\circ$ with the vertical.

This angle is incredibly important. When the beaker is filled with liquid, the light emerging into the air must still travel along this $45^\circ$ path. Therefore, the **angle of emergence** (or angle of incidence in the air, depending on how you trace the ray) is fixed at $i = 45^\circ$.

### The Magic of Refraction

Now, we pour a liquid into the beaker up to a height of $2h$. The problem states that you can now see the *lower end* of the rod.

How is this possible? The light originating from the bottom of the rod at $K(0,0)$ travels through the liquid, hits the liquid-air interface, and refracts (bends). To reach your eye through the pin-hole, it must bend in such a way that it perfectly aligns with our locked $45^\circ$ line of sight.

### Diving into the Geometry of the Liquid

We need to find exactly where the light ray hits the liquid surface. Let's call this point $S$. We know two things about $S$:
1. It lies on the surface of the liquid, so its height is $2h$.
2. It must lie on our locked line of sight, which connects $R(0,h)$ to $P(2h,3h)$.

Since the height of $S$ ($2h$) is exactly halfway between the height of $R$ ($h$) and the height of $P$ ($3h$), by the properties of similar triangles, its horizontal position must also be exactly halfway. The horizontal distance from the left wall is therefore $h$. So, the coordinates of the refraction point are $S(h, 2h)$.

Now we can trace the ray inside the liquid. It travels from the bottom of the rod $K(0,0)$ to the surface point $S(h, 2h)$. Let's find the angle $r$ this ray makes with the vertical normal inside the liquid.

Using the right-angled triangle formed by the ray in the liquid, the horizontal side is $h$ and the vertical side is $2h$. The hypotenuse is:

$$ 	ext{Hypotenuse} = \sqrt{h^2 + (2h)^2} = \sqrt{h^2 + 4h^2} = h\sqrt{5} $$

The sine of the angle $r$ is the opposite side divided by the hypotenuse:

$$ \sin r = \frac{h}{h\sqrt{5}} = \frac{1}{\sqrt{5}} $$

### The Master Equation

We now have all the pieces of the puzzle. We know the angle of the ray in the liquid ($r$) and the angle of the ray in the air ($i = 45^\circ$). It's time to invoke **Snell's Law** at the liquid-air interface:

$$ \mu \sin r = \mu_{	ext{air}} \sin i $$

Substituting the values we've meticulously calculated:

$$ \mu \left(\frac{1}{\sqrt{5}}ight) = 1 \cdot \sin 45^\circ $$

### Final Calculation

We know that $\sin 45^\circ = \frac{1}{\sqrt{2}}$. Plugging this in gives us:

$$ \mu \left(\frac{1}{\sqrt{5}}ight) = \frac{1}{\sqrt{2}} $$

To isolate the refractive index $\mu$, we simply multiply both sides by $\sqrt{5}$:

$$ \mu = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{\frac{5}{2}} $$

And there we have it! By carefully tracking the geometry dictated by a single, tiny pin-hole, we've successfully deduced the optical properties of the mysterious liquid.

Select Answer:

Visualized Solution

\text{Initial Setup and Line of Sight}

\text{Angle of Line of Sight}

\text{Adding the Liquid}

\text{Geometry of the Refracted Ray}

\text{Angle of Incidence in Liquid}

\text{Applying Snell's Law}

\text{Calculating Refractive Index}

The Sigma Insight: Refraction and Total Internal Reflection

The Power of a Fixed Line of Sight

Analyzing the Initial Geometry

The Magic of Refraction

Diving into the Geometry of the Liquid

The Master Equation

Final Calculation

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