An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of

Q: Why does the angle of incidence remain $45^\circ$ when the liquid is added?

The observer's eye and the hole are fixed in place. Therefore, the line of sight from the eye through the hole does not change its angle, regardless of what is inside the jar.

Q: How do we know the horizontal distance is $30\text{ cm}$?

The problem states the jar has a radius of $15\text{ cm}$. The diameter is twice the radius, which gives a total horizontal width of $30\text{ cm}$.

Q: Why does the refracted ray hit the bottom edge?

The problem explicitly states that when the liquid is filled to $30\text{ cm}$, the observer can see the "edge at the bottom of the jar". This means the refracted ray must perfectly strike the corner $(30, 0)$.

Q: Can we use the critical angle concept here?

No, this is a direct application of Snell's Law for refraction from air into a denser medium. Total internal reflection only occurs when light travels from a denser to a rarer medium.

Q: Why do we multiply by 100 at the end?

The problem defines the refractive index as $N/100$. To find the integer value of $N$, we simply multiply our calculated refractive index by 100.

Question

An observer can see through a small hole on the side of a jar (radius $15	ext{ cm}$) at a point at height of $15	ext{ cm}$ from the bottom (see figure). The hole is at a height of $45	ext{ cm}$. When the jar is filled with a liquid up to a height of $30	ext{ cm}$, the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is $N/100$, where $N$ is an integer, the value of $N$ is .......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the observer looking through the small hole in the jar. The jar has a radius of $15	ext{ cm}$, which means its total diameter is $30	ext{ cm}$. The hole you are looking through is positioned at a height of $45	ext{ cm}$ from the bottom.

Initially, when the jar is completely empty, your line of sight travels in a perfectly straight line and hits the opposite wall at a height of $15	ext{ cm}$. Let's break down this geometry. The vertical drop from the hole to the point on the opposite wall is $45	ext{ cm} - 15	ext{ cm} = 30	ext{ cm}$. The horizontal distance across the jar is also $30	ext{ cm}$.

Because the vertical drop and the horizontal distance are exactly equal, the line of sight forms a $45^\circ$ angle with the vertical. This angle is crucial because it will become our angle of incidence when we introduce the liquid.

### The Magic of Refraction

Now, let's pour the liquid into the jar until it reaches a height of $30	ext{ cm}$. Your eye and the hole haven't moved, so your line of sight still enters the jar at that same $45^\circ$ angle.

The ray of light travels through the air and hits the liquid surface. The vertical distance from the hole to the liquid surface is $45	ext{ cm} - 30	ext{ cm} = 15	ext{ cm}$. Since the ray is traveling at a $45^\circ$ angle, the horizontal distance it covers before hitting the liquid is also $15	ext{ cm}$. This means the ray strikes the liquid surface exactly in the middle of the jar!

Once the ray enters the liquid, it refracts (bends) and travels directly to the bottom edge of the jar. The bottom edge is at a horizontal distance of $30	ext{ cm}$ from the hole and a height of $0	ext{ cm}$.

From the point of incidence at the liquid surface, the refracted ray must cover a remaining horizontal distance of $30	ext{ cm} - 15	ext{ cm} = 15	ext{ cm}$, and a vertical distance of $30	ext{ cm} - 0	ext{ cm} = 30	ext{ cm}$.

Let's call the angle of refraction $r$. We can find its tangent:
$$	an r = \frac{	ext{Horizontal Distance}}{	ext{Vertical Distance}} = \frac{15}{30} = \frac{1}{2}$$

Using a simple right triangle where the opposite side is $1$ and the adjacent side is $2$, the hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$. Therefore, the sine of the angle of refraction is:
$$\sin r = \frac{1}{\sqrt{5}}$$

### The Master Equation: Snell's Law

Now we have everything we need to apply Snell's Law at the air-liquid interface. Snell's Law states that the product of the refractive index and the sine of the angle is constant:
$$n_1 \sin i = n_2 \sin r$$

Here, $n_1$ is the refractive index of air (which is $1$), $i$ is the angle of incidence ($45^\circ$), $n_2$ is the refractive index of the liquid (let's call it $\mu$), and $r$ is the angle of refraction.

Substituting our values:
$$1 \cdot \sin 45^\circ = \mu \cdot \frac{1}{\sqrt{5}}$$

$$\frac{1}{\sqrt{2}} = \frac{\mu}{\sqrt{5}}$$

Solving for $\mu$:
$$\mu = \sqrt{\frac{5}{2}} = \sqrt{2.5}$$

### Final Calculation

We found that the refractive index $\mu$ is $\sqrt{2.5}$, which is approximately $1.5811$.

The problem states that the refractive index of the liquid is given by the expression $\frac{N}{100}$, where $N$ is an integer.

Setting our calculated value equal to this expression:
$$\frac{N}{100} = 1.5811$$

$$N = 158.11$$

Since the problem specifies that $N$ must be an integer, we round to the nearest whole number.

**Therefore, the value of $N$ is $158$.**

Enter Numerical Value:

Visualized Solution

Initial Setup and Line of Sight

Angle of Incidence

Adding the Liquid

Angle of Refraction

Calculating $sin r$

Applying Snell's Law

Final Calculation

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Magic of Refraction

The Master Equation

Final Calculation

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Visualized Solution

Initial Setup and Line of Sight

Angle of Incidence

Adding the Liquid

Angle of Refraction

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

Applying Snell's Law

Final Calculation

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Setup

The Magic of Refraction

The Master Equation

Final Calculation

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