A paperweight in the form of a hemisphere of radius 3.0 cm is used to hold down a printed page. An observer looks at the page vertically through the paperweight. At what height above the page will the printed letters near the centre appear to the observer ?

Explanation

To solve the problem, we need to determine the height above the page at which the printed letters appear to the observer when viewed through a hemispherical paperweight.

### Step-by-Step Solution:

1. **Identify Given Data**:
- The radius of the hemisphere (R) = 3.0 cm.
- The observer looks at the page vertically through the paperweight.

2. **Understand the Refraction Process**:
- The light from the printed letters travels through two interfaces:
1. The flat surface of the hemisphere (air to glass).
2. The curved surface of the hemisphere (glass to air).

3. **Calculate Image Distance for the Flat Surface**:
- For the flat surface:
- Refractive index of air (μ1) = 1
- Refractive index of glass (μ2) = 1.5
- Object distance (U) = 0 (since the object is at the center of the hemisphere).
- Radius of curvature (R) = ∞ for a flat surface.
- Using the formula:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
- Substituting the values:
\[ \frac{1.5}{V} - \frac{1}{0} = \frac{1.5 - 1}{\infty} \]
- Since \(U = 0\), the left side becomes undefined, but we can conclude that the image distance (V) for the flat surface is at the same point, hence \(V = 0\).

4. **Calculate Image Distance for the Curved Surface**:
- For the curved surface:
- Refractive index of glass (μ1) = 1.5
- Refractive index of air (μ2) = 1
- Object distance (U) = -3 cm (measured from the pole of the curved surface).
- Radius of curvature (R) = -3 cm (since it is a concave surface).
- Using the same formula:
\[ \frac{1}{V} - \frac{1.5}{-3} = \frac{1 - 1.5}{-3} \]
- Rearranging gives:
\[ \frac{1}{V} + 0.5 = \frac{-0.5}{-3} \]
- Simplifying:
\[ \frac{1}{V} + 0.5 = \frac{1}{6} \]
- Thus:
\[ \frac{1}{V} = \frac{1}{6} - 0.5 = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \]
- Therefore:
\[ V = -3 \text{ cm} \]

5. **Conclusion**:
- The image is formed at the same position as the object, which means the printed letters appear to be at the same height above the page, which is 3 cm.

### Final Answer:
The printed letters near the center appear to the observer at a height of 3.0 cm above the page.

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