A large square container with thin transparent vertical walls and filled with water (refractive inde

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

Since the observer is looking directly from the corner along the diagonal, the rays that enter the observer's eye must emerge from the walls infinitesimally close to the corner. The walls of a square make a $45^\circ$ angle with the diagonal, so the incident rays traveling along the diagonal strike the walls at $45^\circ$.

Q: Why can't we just use the standard apparent depth formula $d' = d/n$?

The standard formula $d' = d/n$ is only valid for near-normal viewing (paraxial rays perpendicular to the surface). Here, the rays strike the surface at a steep angle of $45^\circ$. For oblique viewing, the beam width changes upon refraction, leading to the modified formula $v = u \frac{\cos^2 \theta}{n \cos^2 \alpha}$ for the tangential image.

Q: What does 'tangential image' mean here?

When a wide beam of light refracts obliquely, it doesn't focus at a single point, creating astigmatism. Rays in the plane of incidence form the 'tangential' image, while rays in the perpendicular plane form the 'sagittal' image. For a vertical wire viewed horizontally, the horizontal rays (tangential plane) determine the sharp vertical image the eye focuses on.

Q: Why was this question awarded a BONUS in the actual exam?

The problem statement and the provided diagram had a slight contradiction. The text implies the observer is at the corner near the wire, while the diagram shows the eye far away on the opposite side. This ambiguity in the observer's exact position and pupil size led to it being declared a bonus.

Q: How did we get the angle $\theta - \alpha$ for the image position?

The normal to the wall makes an angle $\alpha$ ($45^\circ$) with the diagonal. The refracted ray makes an angle $\theta$ with the normal. Therefore, the angle between the refracted ray (and thus the image line) and the diagonal is simply the difference, $\theta - \alpha$.

Question

A large square container with thin transparent vertical walls and filled with water (refractive index 4/3) is kept on a horizontal table. A student holds a thin straight wire vertically inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the wire from this corner, another student sees two images of the wire, located symmetrically on each side of the line of sight as shown. The separation (in cm) between these images is_________.

Sigma Percentile · Accepted Answer

### The Setup: A Deceptive Corner

Imagine you are looking at a fish tank, but instead of looking through the flat glass, you press your eye right up against the sharp corner. The world inside distorts, splits, and bends in fascinating ways. This JEE Advanced problem captures exactly that phenomenon.

We have a vertical wire placed $12 	ext{ cm}$ away from the corner of a square container filled with water. When an observer looks at the wire from that very corner, they don't see one wire—they see two! Our mission is to find the exact separation between these twin phantom wires.

### The Geometry of the Line of Sight

First, we must establish the path of the light. The observer is stationed at the corner, looking inwards along the diagonal. For light from the wire to reach the observer's eye, it must travel through the water and strike the glass walls infinitesimally close to the corner.

Because the container is a square, the diagonal perfectly bisects the $90^\circ$ corner. This means the walls make an angle of $45^\circ$ with the diagonal. Consequently, the light rays traveling from the wire along the diagonal will strike the walls with an angle of incidence $\alpha = 45^\circ$.

### The Law of Refraction

Light hates traveling in straight lines when it crosses boundaries. As the rays hit the water-air interface, they bend. We invoke Snell's Law to find the angle of refraction, $	heta$:

$$n \sin \alpha = 1 \cdot \sin 	heta$$

Substituting the refractive index of water ($n = 4/3$) and our angle of incidence:

$$\frac{4}{3} \sin 45^\circ = \sin 	heta$$

$$\sin 	heta = \frac{4}{3} \cdot \frac{1}{\sqrt{2}} = \frac{2\sqrt{2}}{3}$$

Using the fundamental trigonometric identity, we can also find $\cos 	heta$:

$$\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - \frac{8}{9}} = \frac{1}{3}$$

### The Secret Formula of Oblique Viewing

Here is where most students fall into a trap. The standard apparent depth formula, $d' = d/n$, is strictly for paraxial rays—rays that hit the surface almost perpendicularly. But our rays are hitting at a steep $45^\circ$ angle!

When viewing obliquely, the bundle of light rays compresses or expands upon refraction. This astigmatic effect means the image forms at a different distance. For rays in the plane of incidence (which form the sharp vertical image of the wire), the apparent distance $x$ from the point of incidence is given by:

$$x = u \frac{\cos^2 	heta}{n \cos^2 \alpha}$$

Let's plug in our hard-earned values. The actual distance $u$ is $12 	ext{ cm}$:

$$x = 12 \cdot \frac{(1/3)^2}{(4/3) \cdot (1/\sqrt{2})^2}$$

Simplifying the fractions:

$$x = 12 \cdot \frac{1/9}{(4/3) \cdot (1/2)} = 12 \cdot \frac{1/9}{2/3} = 12 \cdot \frac{1}{6} = 2 	ext{ cm}$$

The image is pulled drastically closer, appearing just $2 	ext{ cm}$ away from the corner!

### The Final Separation

We aren't done yet. The container has two walls forming the corner, so the light splits, creating two symmetric images.

Where exactly are these images? The refracted ray makes an angle $	heta$ with the normal. The diagonal makes an angle $\alpha$ with the normal. Therefore, the angle between the backward-extended refracted ray (where the image lies) and the diagonal is simply $	heta - \alpha$.

The perpendicular distance from one image to the diagonal is $x \sin(	heta - \alpha)$. Since there are two images, the total separation $d$ is double that:

$$d = 2x \sin(	heta - \alpha)$$

Using the sine subtraction formula:

$$d = 2(2) [\sin 	heta \cos \alpha - \cos 	heta \sin \alpha]$$

Substitute the trigonometric values:

$$d = 4 \left[ \frac{2\sqrt{2}}{3} \cdot \frac{1}{\sqrt{2}} - \frac{1}{3} \cdot \frac{1}{\sqrt{2}} ight]$$

$$d = 4 \left[ \frac{2}{3} - \frac{\sqrt{2}}{6} ight] = \frac{8 - 2\sqrt{2}}{3} 	ext{ cm}$$

This evaluates to approximately $1.73 	ext{ cm}$.

### The "Bonus" Mystery

You might wonder why this beautifully complex question was awarded a **BONUS** in the official JEE Advanced grading. The devil is in the details. The problem statement implies the observer is at the corner near the wire, but the provided diagram schematically placed the eye far away on the opposite side. This ambiguity regarding the observer's exact position and the pupil's acceptance angle made the strict mathematical interpretation debatable. Nevertheless, the physics of oblique viewing it teaches remains an absolute masterclass!

Visualized Solution

$u = 12 cm$

$α = 4 5^{\circ}$

$n sin α = 1 \cdot sin θ$

$sin θ = \frac{2 2}{3}$

$x = u \frac{c o s ^{2} θ}{n c o s ^{2} α}$

$x = 2 cm$

\text{Angle} = \theta - \alpha

$d = 2 x sin (θ - α)$

$d \approx 1.73 cm$

\text{What if } \alpha = 30^\circ?

The Sigma Insight: Refraction and Total Internal Reflection

The Setup

The Geometry of the Line of Sight

The Law of Refraction

The Secret Formula of Oblique Viewing

The Final Separation

The "Bonus" Mystery

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