A coin is placed at the bottom of a beaker containing water (refractive index=4/3) at a depth of 12cm. By what height the coin appears to be raised when seen from vertical above ?

Answer

To solve the problem of how much the coin appears to be raised when viewed from above the water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Actual depth of the coin (dA) = 12 cm
- Refractive index of water (n) = 4/3

2. **Use the Formula for Apparent Depth:**
The formula for apparent depth (dA') when light passes from a medium of higher refractive index to a medium of lower refractive index is given by:
\[ dA' = \frac{dA}{n} \]
Substituting the known values:
\[ dA' = \frac{12 \text{ cm}}{4/3} \]

3. **Calculate the Apparent Depth:**
To perform the division, we can multiply by the reciprocal of the refractive index:
\[ dA' = 12 \text{ cm} \times \frac{3}{4} = 9 \text{ cm} \]

4. **Determine the Shift in Position:**
The shift (s) in the position of the coin can be calculated using the formula:
\[ s = dA - dA' \]
Substituting the values we have:
\[ s = 12 \text{ cm} - 9 \text{ cm} = 3 \text{ cm} \]

5. **Conclusion:**
The coin appears to be raised by 3 cm when viewed from directly above the water.

### Final Answer:
The coin appears to be raised by **3 cm**.

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