An image is formed at a distance of 100 cm from the glass surface with refractive index 1.5, when a point object is placed in the air at a distance of 100 cm from the glass surface. The radius of curvature is of the surface is

Options

A. 20 cm
B. 40 cm
C. 30 cm
D. 50 cm

Answer

To solve the problem step by step, we will use the lens maker's formula and the given information about the object, image, and refractive indices.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Refractive index of air (μ1) = 1
- Refractive index of glass (μ2) = 1.5
- Object distance (u) = -100 cm (negative because the object is on the same side as the incoming light)
- Image distance (v) = +100 cm (positive because the image is formed on the opposite side of the incoming light)

2. **Use the Lens Maker's Formula:**
The formula relating the refractive indices and distances is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
where \( R \) is the radius of curvature.

3. **Substitute the Known Values:**
Substitute the values into the formula:
\[ \frac{1.5}{100} - \frac{1}{-100} = \frac{1.5 - 1}{R} \]

4. **Simplify the Left Side:**
Calculate the left side:
\[ \frac{1.5}{100} + \frac{1}{100} = \frac{1.5 + 1}{100} = \frac{2.5}{100} \]
So, we have:
\[ \frac{2.5}{100} = \frac{0.5}{R} \]

5. **Cross-Multiply to Solve for R:**
Cross-multiplying gives:
\[ 2.5R = 0.5 \times 100 \]
\[ 2.5R = 50 \]

6. **Calculate R:**
Now, divide both sides by 2.5:
\[ R = \frac{50}{2.5} = 20 \text{ cm} \]

### Final Answer:
The radius of curvature \( R \) of the glass surface is **20 cm**.