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 ray from the eye to the liquid surface remain the same?

The observer's eye is fixed at the hole, and they are looking along a straight line towards the liquid surface. Until the light hits the liquid, it travels in air without bending, so its path is identical to the empty jar scenario.

Q: How do we know the horizontal distance the ray travels in the liquid is 15 cm?

The total width of the jar is 30 cm. The ray travels 15 cm horizontally in the air before hitting the liquid surface. Therefore, the remaining horizontal distance to the opposite bottom edge must be 30 cm - 15 cm = 15 cm.

Q: Why is the angle of incidence $45^\circ$?

The ray drops 15 cm vertically and travels 15 cm horizontally. This forms a right-angled triangle with equal opposite and adjacent sides. The tangent of the angle is 1, which corresponds to $45^\circ$.

Q: Can we use the critical angle concept here?

No, this is a direct application of Snell's Law for refraction from a rarer medium (air) to a denser medium (liquid). Total internal reflection and critical angles only apply when light travels from a denser to a rarer medium.

Q: Why do we round $158.1$ to $158$?

The problem explicitly states that $N$ is an integer. Since $158.1$ is closest to $158$, we round it down to satisfy the integer constraint.

Question

An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 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

The problem of the refracting jar is a beautiful interplay between simple geometry and the fundamental laws of optics. It challenges us to trace the path of light as it travels through different mediums, bending and revealing hidden corners. Let's dive into the step-by-step breakdown of this fascinating phenomenon.

### Analyzing the Empty Jar

Imagine you are looking through a small hole on the side of an empty jar. The hole is located at a height of $45	ext{ cm}$. When you look straight through, your line of sight hits the opposite wall at a height of $15	ext{ cm}$.

Since light travels in a straight line in a uniform medium (air), we can determine the slope of your line of sight. The vertical drop is $45	ext{ cm} - 15	ext{ cm} = 30	ext{ cm}$. The jar has a radius of $15	ext{ cm}$, which means its total width (diameter) is $30	ext{ cm}$.

Therefore, the light ray drops $30	ext{ cm}$ vertically over a horizontal distance of $30	ext{ cm}$. This perfect $1:1$ ratio means the ray travels at a $45^\circ$ angle relative to the horizontal and vertical axes.

### The Filled Jar and the Angle of Incidence

Now, the jar is filled with a liquid up to a height of $30	ext{ cm}$. The light ray from your eye to the liquid surface remains completely unchanged because it is still traveling through the air.

The ray starts at $y = 45	ext{ cm}$ and hits the liquid surface at $y = 30	ext{ cm}$. This is a vertical drop of $15	ext{ cm}$. Because the slope of the ray is $1$, it must also travel $15	ext{ cm}$ horizontally.

If we draw a normal (a vertical line) at the point where the ray hits the liquid, we can find the angle of incidence, $i$.

$$ 	an i = \frac{15}{15} = 1 \implies i = 45^\circ $$

This gives us $\sin i = \frac{1}{\sqrt{2}}$.

### Tracing the Refracted Ray

Once the ray enters the liquid, it bends due to refraction. The problem states that you can now see the edge at the bottom of the jar. This means the refracted ray travels from the point of incidence on the surface down to the opposite bottom corner.

Let's look at the geometry of this refracted ray. It starts at a horizontal distance of $15	ext{ cm}$ and ends at the opposite wall, which is at $30	ext{ cm}$. So, it travels $15	ext{ cm}$ horizontally inside the liquid. Vertically, it travels from the surface at $30	ext{ cm}$ down to the bottom at $0	ext{ cm}$, covering a distance of $30	ext{ cm}$.

We can now find the angle of refraction, $r$, using this new right-angled triangle.

$$ 	an r = \frac{15}{30} = \frac{1}{2} $$

Using the Pythagorean theorem, the hypotenuse of this triangle is $\sqrt{15^2 + 30^2} = 15\sqrt{5}$. Therefore, the sine of the angle of refraction is:

$$ \sin r = \frac{15}{15\sqrt{5}} = \frac{1}{\sqrt{5}} $$

### The Master Equation: Snell's Law

With both the angle of incidence and the angle of refraction known, we can apply Snell's Law at the air-liquid interface to find the refractive index of the liquid, $\mu$.

$$ n_{	ext{air}} \sin i = n_{	ext{liquid}} \sin r $$

Substituting our known values ($n_{	ext{air}} = 1$):

$$ 1 \cdot \sin 45^\circ = \mu \cdot \frac{1}{\sqrt{5}} $$

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

### Final Calculation

Now, it's just a matter of simple algebra to isolate $\mu$:

$$ \mu = \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{2.5} \approx 1.581 $$

The problem states that the refractive index is given by the expression $\frac{N}{100}$. We can set up our final equation:

$$ \frac{N}{100} = 1.581 $$

$$ N = 158.1 $$

Since the question asks for an integer value for $N$, we round $158.1$ to the nearest whole number.

**Final Answer:** The value of $N$ is **$158$**.

Enter Numerical Value:

Visualized Solution

$I ni t ia l O b ser v a t i o n (E m pt y J a r)$

$A n g l e o f I n c i d e n ce$

$R e f r a c t i o n an d t h e F ina l R a y$

$A n g l e o f R e f r a c t i o n$

$A ppl y in g S n e l l^{'} s L a w$

$C a l c u l a t in g R e f r a c t i v e I n d e x$

The Sigma Insight: Refraction and Total Internal Reflection

Analyzing the Empty Jar

The Filled Jar and the Angle of Incidence

Tracing the Refracted Ray

The Master Equation

Final Calculation

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