A container is filled with water (μ = 1.33) up to a height of 33.25 cm. A concave mirror is placed 1

Q: Why do we calculate the apparent depth?

Because the light rays from the object travel from water (denser) to air (rarer) before reaching the mirror. This refraction causes the object to appear closer to the surface than it actually is.

Q: Why did we ignore the second refraction when the light re-enters the water?

In many classic physics problems, the phrasing 'image is formed at...' refers to the image produced by the primary optical element (the mirror) unless specified otherwise. If we accounted for the second refraction, the focal length would be 18.3 cm, which isn't an option. The exact option 20 cm confirms our assumption.

Q: What does it mean when $u = v$?

When the object distance and image distance are equal in a spherical mirror, it means the object is placed exactly at the center of curvature. At this point, the rays hit the mirror normally and retrace their paths.

Q: How do we know the image is real?

Since the image is formed in front of the concave mirror (below it, in this setup), the light rays actually converge at that point. Therefore, the image is real and $v$ is negative according to the sign convention.

Q: Can we use the formula $d_{app} = d_{real} / \mu$ for any viewing angle?

Strictly speaking, this formula is valid only for near-normal viewing (paraxial rays). In standard optics problems like this, we always assume paraxial rays unless the viewing angle is explicitly given.

Question

A container is filled with water ($\mu = 1.33$) up to a height of $33.25	ext{ cm}$. A concave mirror is placed $15	ext{ cm}$ above the water level and the image of an object placed at the bottom is formed $25	ext{ cm}$ below the water level. The focal length of the mirror is

Options:

(A) $10	ext{ cm}$

(B) $15	ext{ cm}$

(C) $20	ext{ cm}$

(D) $25	ext{ cm}$

Sigma Percentile · Accepted Answer

This problem is a beautiful interplay between refraction at a plane surface and reflection from a spherical mirror. It also contains a subtle trap in its phrasing that we must navigate carefully. Let's break it down step by step.

### Analyzing the Setup

Imagine you are a light ray starting at the bottom of a container filled with water. The depth of the water is $33.25	ext{ cm}$, and the refractive index is $\mu = 1.33$ (which is exactly $4/3$). High above the water, at a distance of $15	ext{ cm}$ from the surface, a concave mirror is waiting to reflect you back.

### The Apparent Depth

When light rays from the object travel upwards and hit the water-air interface, they move from a denser medium (water) to a rarer medium (air). According to Snell's Law, they bend away from the normal.

To an observer in the air—or in this case, the concave mirror—the object appears to be lifted up. This is the concept of **apparent depth**. We can calculate it using the standard formula:

$$d_{	ext{app}} = \frac{d_{	ext{real}}}{\mu}$$

Substituting our values:

$$d_{	ext{app}} = \frac{33.25}{4/3} = 25	ext{ cm}$$

So, the mirror doesn't see the object at the bottom; it sees a virtual object $O'$ located $25	ext{ cm}$ below the water surface.

### The Mirror's Perspective

Now, let's determine the object distance ($u$) for the concave mirror. The mirror is $15	ext{ cm}$ above the water, and the apparent object is $25	ext{ cm}$ below the water.

The total distance from the mirror to the apparent object is:

$$15	ext{ cm} + 25	ext{ cm} = 40	ext{ cm}$$

Using the standard Cartesian sign convention (where the direction of incident light is positive), the object distance is:

$$u = -40	ext{ cm}$$

### The Subtle Trap

The problem states: *"the image of an object placed at the bottom is formed $25	ext{ cm}$ below the water level."*

Here lies the ambiguity. Does this refer to the image formed directly by the mirror, or the final image after the light refracts back into the water?

In many classic physics problems of this type, unless specified otherwise, the position given refers to the image produced by the primary optical element (the mirror). Let's assume the image formed by the mirror is at $25	ext{ cm}$ below the water level.

This means the image distance ($v$) from the mirror is:

$$v = -(15	ext{ cm} + 25	ext{ cm}) = -40	ext{ cm}$$

### Final Calculation

Look closely at our values:

$$u = -40	ext{ cm}$$
$$v = -40	ext{ cm}$$

The image distance is exactly equal to the object distance! When does a concave mirror form an image at the exact same location as the object? This happens only when the object is placed at the **center of curvature** of the mirror.

Therefore, the radius of curvature ($R$) is $40	ext{ cm}$. The focal length ($f$) is half of the radius of curvature:

$$f = \frac{R}{2} = \frac{40}{2} = 20	ext{ cm}$$

This matches option (c) perfectly.

*(Note: If we had assumed the $25	ext{ cm}$ was the final image after the second refraction, the mirror would have had to form its image at $18.75	ext{ cm}$ below the water, leading to a focal length of $\approx 18.3	ext{ cm}$. Since $20	ext{ cm}$ is an exact option, our assumption aligns perfectly with the examiner's intent!)*

Animated Solution for Physics - Optics: A container is filled with water ( $μ = 1.33$ ) up to a height of $33.25 cm$ . A concave mirror is placed $15 cm$ above the water level and the image of an object placed at the bottom is formed $25 cm$ below the water level. The focal length of the mirror is

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Visualized Solution

\text{Analyzing the Setup}

\text{Apparent Depth of the Object}

\text{Object Distance for the Mirror}

\text{Image Distance}

\text{Focal Length Calculation}

\text{The Way Forward: The Second Refraction}

The Sigma Insight: Spherical Mirror

Analyzing the Setup

The Apparent Depth

The Mirror's Perspective

The Subtle Trap

Final Calculation

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