1. Given data

• Concave mirror radius of curvature: $R = 40\,\text{cm}$
• Hence focal length: $f = \frac{R}{2} = 20\,\text{cm}$
• Water depth above mirror: $5\,\text{cm}$
• Refractive index of water: $\mu = 1.33$
• Object (small particle) is floating on the water surface.

We need the final image position as seen from directly above the water surface.

2. Image formed by the concave mirror inside water

The particle is at the water surface, and the mirror is $5\,\text{cm}$ below it. So the object distance for the mirror is $u = 5\,\text{cm}$

For a concave mirror, using the mirror formula: $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Substitute $f=20\,\text{cm}$ and $u=5\,\text{cm}$: $\frac{1}{20} = \frac{1}{5} + \frac{1}{v}$

So, $\frac{1}{v} = \frac{1}{20} - \frac{1}{5} = \frac{1-4}{20} = -\frac{3}{20}$

Thus, $v = -\frac{20}{3} \approx -6.67\,\text{cm}$

The negative sign means the image is formed behind the mirror.

Since the mirror is at depth $5\,\text{cm}$ below the surface, this mirror image is at a depth below the surface of $5 + 6.67 = 11.67\,\text{cm}$

So, for refraction at the water surface, there is an object inside water at real depth $h = 11.67\,\text{cm}$

3. Apparent depth when seen from air

An object inside water at real depth $h$ appears from air at apparent depth $h' = \frac{h}{\mu}$

Therefore, $h' = \frac{11.67}{1.33} \approx 8.77\,\text{cm}$

Hence, $d \approx 8.8\,\text{cm}$

4. Match with options

The closest option is:

• A: $11.7\,\text{cm}$
• B: $6.7\,\text{cm}$
• C: $13.4\,\text{cm}$
• D: $8.8\,\text{cm}$

So the correct answer is: $\boxed{\text{D }\; 8.8\,\text{cm}}$

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.