A plano convex lens `(mu=1.5)` has a maximum thickness of `1mm` .If diameter of its aperture `4cm` Find
(i)Radius of curvature of curved surface
(ii) its focal length in air.

Answer

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Given:
- Refractive index of the lens, \( \mu = 1.5 \)
- Maximum thickness of the lens, \( t = 1 \text{ mm} = 0.1 \text{ cm} \)
- Diameter of the aperture, \( D = 4 \text{ cm} \) (which gives a radius of aperture \( r_a = 2 \text{ cm} \))

### (i) Finding the Radius of Curvature of the Curved Surface

1. **Understanding the Geometry**:
- The lens is plano-convex, meaning one side is flat (plane) and the other side is curved.
- The maximum thickness occurs at the center of the lens.

2. **Using the Pythagorean Theorem**:
- For a plano-convex lens, we can relate the radius of curvature \( R \) to the thickness \( t \) and the radius of the aperture \( r_a \).
- According to the Pythagorean theorem:
\[ R^2 = r_a^2 + (R - t)^2 \]
- Substituting the values:
\[ R^2 = (2 \text{ cm})^2 + (R - 0.1 \text{ cm})^2 \]
- Expanding the equation:
\[ R^2 = 4 + (R^2 - 0.2R + 0.01) \]
- Simplifying:
\[ 0 = 4 + 0.01 - 0.2R \]
\[ 0.2R = 4.01 \]
\[ R = \frac{4.01}{0.2} = 20.05 \text{ cm} \approx 20 \text{ cm} \]

### (ii) Finding the Focal Length in Air

1. **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 a plano-convex lens:
- \( R_1 = R = 20 \text{ cm} \) (convex surface)
- \( R_2 = \infty \) (plane surface)
- Substituting the values into the formula:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - 0 \right) \]
\[ \frac{1}{f} = 0.5 \cdot \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40} \]
- Therefore, the focal length \( f \) is:
\[ f = 40 \text{ cm} \]

### Final Answers:
- (i) Radius of curvature \( R \approx 20 \text{ cm} \)
- (ii) Focal length \( f = 40 \text{ cm} \)