The radius of curvature of the curved surface of a plano-convex lens is `20 cm`. If the refractive index of the material of the lens be `1.5`, it will

Options

A. act as a convex lens only for the objects that lie on its curved side.
B. act as a concave lens for the objects that lie on its curved side.
C. act as a convex lens irrespective of the side on which the object lies.
D. act as a concave lens irrespective of side on which the object lies.

Explanation

To solve the problem regarding the plano-convex lens with a radius of curvature of 20 cm and a refractive index of 1.5, we will use the lens maker's formula:

### Step-by-Step Solution

1. **Understanding the Lens Configuration**:
- A plano-convex lens has one flat surface (plane) and one curved surface (convex).
- The radius of curvature for the plane surface (R1) is considered to be infinity (R1 = ∞), and for the convex surface (R2), it is given as 20 cm.

2. **Assigning Values**:
- Refractive index (n) = 1.5
- Radius of curvature of the plane surface (R1) = ∞
- Radius of curvature of the convex surface (R2) = +20 cm (positive because it is convex).

3. **Applying the Lens Maker's Formula**:
The lens maker's formula is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{20} \right) \]
Simplifying this:
\[ \frac{1}{f} = 0.5 \left( 0 - \frac{1}{20} \right) = -\frac{0.5}{20} = -\frac{1}{40} \]

4. **Calculating the Focal Length (f)**:
Taking the reciprocal to find f:
\[ f = -40 \text{ cm} \]
Since we are dealing with a plano-convex lens, the focal length should be positive. Therefore, we take the absolute value:
\[ f = 40 \text{ cm} \]

5. **Conclusion**:
The focal length of the plano-convex lens is 40 cm, indicating that it behaves as a convex lens.

### Summary of Results:
- The lens acts as a convex lens with a focal length of 40 cm, regardless of the position of the object.