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 of finding the height above the page at which the printed letters appear to the observer when looking through a hemispherical paperweight, we can follow these steps:

### Step 1: Understand the Geometry
The paperweight is a hemisphere with a radius of 3.0 cm. The observer is looking vertically down through the paperweight at a printed page.

### Step 2: Identify the Object and Image
The printed letters can be considered as an object located at the flat surface of the hemisphere. The observer will see an image of these letters after refraction through the curved surface of the hemisphere.

### Step 3: Refraction at the Flat Surface
For the flat surface of the hemisphere, we can use the formula for refraction:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- \(\mu_1\) = refractive index of air = 1
- \(\mu_2\) = refractive index of glass = 1.5
- \(R\) = radius of curvature of the flat surface = ∞ (infinity)
- \(U\) = object distance = 0 (the object is at the flat surface)

Substituting these values into the equation:
\[ \frac{1.5}{V} - \frac{1}{0} = \frac{1.5 - 1}{\infty} \]
Since the term with \(U\) approaches infinity, we can conclude that the image is formed at the flat surface, which means \(V = 0\).

### Step 4: Refraction at the Curved Surface
Next, we consider the refraction at the curved surface of the hemisphere. We will use the same formula:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- \(U\) = object distance from the pole of the curved surface = -3 cm (since it is measured from the pole downwards)
- \(R\) = radius of curvature = -3 cm (the radius is negative for the curved surface)
- \(\mu_1 = 1.5\), \(\mu_2 = 1\)

Substituting these values:
\[ \frac{1}{V} - \frac{1.5}{-3} = \frac{1 - 1.5}{-3} \]
This simplifies to:
\[ \frac{1}{V} + 0.5 = \frac{-0.5}{-3} \]
\[ \frac{1}{V} + 0.5 = \frac{1}{6} \]
\[ \frac{1}{V} = \frac{1}{6} - 0.5 = \frac{1}{6} - \frac{3}{6} = -\frac{2}{6} = -\frac{1}{3} \]
Thus, \(V = -3\) cm.

### Step 5: Interpretation of the Result
The negative value of \(V\) indicates that the image is formed 3 cm below the flat surface of the hemisphere, which corresponds to the height above the printed page.

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

---