A container is filled with water (\(\mu =4/3\)) up to a height of 24 cm. A concave mirror is placed 10 cm above the water level, and the image (I) of an object placed at the bottom is formed 20 cm below the water level, as shown in Figure. The focal length of the mirror is nearly
Show Hint
Convert the real depths to apparent positions as seen from air, then apply the mirror formula.
Step 1: Understanding the Concept
Water shifts the apparent position of submerged points. Apparent depth \(=\dfrac{\text{real depth}}{\mu}\). We find the apparent object and image positions and then use the mirror formula.
Step 2: Object distance
The object is 24 cm deep, so its apparent depth is \(\dfrac{24}{4/3}=18\ \text{cm}\) below the surface. The mirror is 10 cm above, so \(u=-(18+10)=-28\ \text{cm}\).
Step 3: Image distance
The final image is 20 cm deep, which appears at \(\dfrac{20}{4/3}=15\ \text{cm}\) below the surface. So \(v=-(15+10)=-25\ \text{cm}\).
Step 4: Mirror formula
\(\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{25}-\dfrac{1}{28}=-\dfrac{53}{700}\), so \(f\approx-13.2\ \text{cm}\). It is negative as expected for a concave mirror.
Final Answer:
The focal length of the mirror is nearly \(-13.2\ \text{cm}\).
\[ \boxed{\text{(A) }-13.2\ \text{cm}} \]
