A glass sphere has radius of `5.0 cm` and a refractive index of `1.6.` A paperweight is constructed by slicing through the sphere on a plate that is `2.0 cm` from the centre of the sphere and perpendicular to radius of the sphere that passes through the centre of the circle formed by the intersection of the plane and the sphere. The paperweight is placed on a table and viewed from directly above an observer who is `8.0 cm` from the table top, as shown in figure. When viewed through the paperweight, how far away does the table top appear to the observer?

Explanation

To solve the problem step by step, we will use the lens maker's formula and the concepts of refraction. Here’s how we can find out how far away the table top appears to the observer:

### Step 1: Identify the parameters
- Radius of the glass sphere (R) = 5.0 cm
- Refractive index of glass (μ1) = 1.6
- Refractive index of air (μ2) = 1.0
- Distance from the center of the sphere to the plane (u) = -3.0 cm (since it is measured from the center of the sphere to the plane)
- Observer's height above the table = 8.0 cm

### Step 2: Calculate the object distance (u)
The object distance (u) is the distance from the object (the table) to the surface of the paperweight. Since the paperweight is 2.0 cm from the center of the sphere, the distance from the top of the sphere to the top of the paperweight is:
\[ u = R - \text{distance from center to plane} = 5.0 \, \text{cm} - 2.0 \, \text{cm} = 3.0 \, \text{cm} \]
However, since we are considering the direction of light travel as positive, we take:
\[ u = -3.0 \, \text{cm} \]

### Step 3: Use the lens maker's formula
Using the formula:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Substituting the values:
\[ \frac{1.0}{v} - \frac{1.6}{-3.0} = \frac{1.0 - 1.6}{-5.0} \]
This simplifies to:
\[ \frac{1.0}{v} + \frac{1.6}{3.0} = \frac{-0.6}{-5.0} \]
Calculating the right side:
\[ \frac{1.0}{v} + \frac{0.5333}{1} = 0.12 \]
So:
\[ \frac{1.0}{v} = 0.12 - 0.5333 = -0.4133 \]
Thus:
\[ v = -2.42 \, \text{cm} \]

### Step 4: Calculate the apparent distance to the observer
The observer is 8.0 cm above the table. The distance from the top of the paperweight to the table is:
\[ \text{Distance from observer to table} = 8.0 \, \text{cm} - 3.0 \, \text{cm} = 5.0 \, \text{cm} \]
Now, we add the image distance (which is negative indicating it is below the table):
\[ \text{Apparent distance} = 5.0 \, \text{cm} + 2.42 \, \text{cm} = 7.42 \, \text{cm} \]

### Final Answer
The table top appears to be **7.42 cm** away from the observer.

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