A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water fil

Q: Why is the object distance $u$ taken as negative?

By standard sign convention, the direction of incident light is taken as positive. If we place the origin at the mirror's pole, the incident light travels downwards. The object (particle) is above the mirror, opposite to the direction of incident light, making $u = -5\text{ cm}$.

Q: What does a positive image distance $v$ mean here?

A positive $v$ means the image is formed in the direction of the incident light. In this setup, since incident light travels downwards, a positive $v$ places the image *below* the mirror. It is a virtual image formed behind the reflecting surface.

Q: Why do we add $5\text{ cm}$ to $v$ for the second part of the problem?

The refraction happens at the water surface, not at the mirror. The image $I_1$ is $\frac{20}{3}\text{ cm}$ below the mirror, and the mirror itself is $5\text{ cm}$ below the surface. Therefore, the total real depth of $I_1$ from the refracting surface is $5 + \frac{20}{3}\text{ cm}$.

Q: Why does the image $I_1$ act as an object for the water surface?

In multiple optical systems, the light rays leaving the first element (mirror) enter the second element (water surface). The point from where these rays appear to diverge (or converge) serves as the object for the subsequent optical event.

Q: Would the apparent depth change if the observer was looking from an angle?

Yes. The formula $d = \frac{h}{\mu}$ is an approximation valid only for near-normal viewing (paraxial rays). If viewed from a significant angle, the apparent depth decreases further due to the geometry of refraction.

Question

A concave mirror has radius of curvature of $40	ext{ cm}$. It is at the bottom of a glass that has water filled up to $5	ext{ cm}$ (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance $d$ from the surface of water. The value of $d$ is close to
[Refractive index of water $= 1.33$]

Options:

(A) $6.7	ext{ cm}$

(B) $13.4	ext{ cm}$

(C) $8.8	ext{ cm}$

(D) $11.7	ext{ cm}$

Sigma Percentile · Accepted Answer

The problem presents a fascinating interplay between reflection and refraction. We have a particle floating on the surface of water, and a concave mirror sitting at the bottom of the glass. To find the final apparent position of the particle's image, we must trace the journey of light rays in two distinct phases: first, their reflection from the concave mirror, and second, their refraction as they exit the water surface.

### Phase 1: Reflection from the Concave Mirror

Light rays originating from the particle travel downwards through the water and strike the concave mirror. To apply the mirror formula, we must establish a clear sign convention. Let's place the origin at the pole of the mirror and take the downward direction (the direction of incident light) as positive.

The particle is located at the water surface, which is $5	ext{ cm}$ above the mirror. Therefore, the object distance is:
$$u = -5	ext{ cm}$$

The radius of curvature of the concave mirror is given as $R = 40	ext{ cm}$. The center of curvature lies above the mirror, so the focal length is:
$$f = -\frac{R}{2} = -20	ext{ cm}$$

Now, we invoke the mirror formula to find the position of the first image, $I_1$:
$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

Substituting our values:
$$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-20}$$

Rearranging to solve for $v$:
$$\frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{4 - 1}{20} = \frac{3}{20}$$

$$v = +\frac{20}{3}	ext{ cm}$$

The positive sign is crucial here. It indicates that the image $I_1$ is formed **behind** the mirror (further down in our coordinate system).

### Phase 2: Refraction at the Water Surface

The rays reflected from the mirror travel back upwards towards the water surface. To an observer looking from above, these diverging rays appear to originate from the image $I_1$. Thus, $I_1$ acts as a real object for the water-air interface.

We need to determine the total depth of this object $I_1$ from the water surface. The mirror is $5	ext{ cm}$ below the surface, and $I_1$ is $\frac{20}{3}	ext{ cm}$ below the mirror. 
Total real depth, $h = 5 + \frac{20}{3} = \frac{15 + 20}{3} = \frac{35}{3}	ext{ cm}$

As the light rays exit the water (a denser medium) into the air (a rarer medium), they bend away from the normal. This refraction causes the image to appear shifted upwards. The apparent depth $d$ is given by the relation:
$$d = \frac{	ext{Real Depth}}{\mu_{	ext{water}}}$$

Substituting the values:
$$d = \frac{35/3}{4/3} = \frac{35}{4} = 8.75	ext{ cm}$$

### Final Conclusion

The final image of the particle appears to be at a depth of **$8.75	ext{ cm}$** from the surface. Looking at the given options, $8.8	ext{ cm}$ is the closest approximation. This elegant problem beautifully demonstrates how multiple optical elements can be analyzed sequentially by treating the image of the first element as the object for the second.

Select Answer:

Visualized Solution

Analyzing the Setup

Reflection from Concave Mirror

Applying the Mirror Formula

Calculating Image Distance $v$

Refraction at Water Surface

Calculating Apparent Depth

Conclusion

The Sigma Insight: Refraction and Total Internal Reflection

Phase 1

Phase 2

Final Conclusion

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