A parallel beam of light falls on the surface of a convex lens whose radius of curvature of both sides of 20cm. The refractive index of the material of the lens varies as `mu=1.5+0.5r` , where r is the distance of the point on the aperture from the optical centre in cm. Find the length of the region on the axis of the lens where the light will appear. The radius of aperture of the lens is 1cm.

Explanation

To solve the problem, we need to find the length of the region on the axis of the lens where the light will appear, given that the refractive index of the lens varies with the distance from the optical center.

### Step-by-Step Solution:

1. **Understanding the Given Parameters:**
- Radius of curvature (R1 and R2) of the lens = 20 cm (both sides).
- Radius of the lens aperture = 1 cm.
- Refractive index formula: \( \mu = 1.5 + 0.5r \), where \( r \) is the distance from the optical center in cm.

2. **Using the Lens Maker's Formula:**
The lens maker's formula is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For our lens, since \( R_1 = +20 \) cm and \( R_2 = -20 \) cm (the convention for the second surface), we can substitute these values into the formula.

3. **Substituting the Refractive Index:**
Substitute \( \mu \) into the lens maker's formula:
\[ \frac{1}{F} = (1.5 + 0.5r - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) \]
This simplifies to:
\[ \frac{1}{F} = (0.5 + 0.5r) \left( \frac{1}{20} + \frac{1}{20} \right) = (0.5 + 0.5r) \left( \frac{2}{20} \right) = (0.5 + 0.5r) \left( \frac{1}{10} \right) \]

4. **Simplifying Further:**
\[ \frac{1}{F} = \frac{0.5 + 0.5r}{10} = \frac{1 + r}{20} \]
Therefore, we can express \( F \) as:
\[ F = \frac{20}{1 + r} \]

5. **Finding the Focal Length for Different Values of r:**
- For \( r = 0 \) cm (center of the lens):
\[ F = \frac{20}{1 + 0} = 20 \text{ cm} \]
- For \( r = 1 \) cm (edge of the lens):
\[ F = \frac{20}{1 + 1} = 10 \text{ cm} \]

6. **Determining the Region on the Axis:**
The light will converge at different focal lengths depending on the position \( r \) from the optical center. The light will appear to focus between the two focal lengths:
- From \( F = 10 \) cm (at \( r = 1 \) cm) to \( F = 20 \) cm (at \( r = 0 \) cm).

7. **Calculating the Length of the Region:**
The length of the region on the axis where the light will appear is:
\[ \text{Length} = 20 \text{ cm} - 10 \text{ cm} = 10 \text{ cm} \]

### Final Answer:
The length of the region on the axis of the lens where the light will appear is **10 cm**.