The radii of curvature of two spherical surfaces of a concave convex lens `(mu=1.5)` are 20 cm and 40 cm. (i) What is its focal length when it is in air? Also find its focal length when it is immersed in a liquid of refractive index (ii) `mu=1.2` (iii) `mu=2`

Options

A. (80cm, 160cm, -160cm)
B. (80cm, 80cm,-80cm)
C. (160cm, 80cm, 80cm)
D. (80cm, 160cm, 160cm)

Answer

To solve the problem, we will use the lens maker's formula to find the focal length of a concave-convex lens in different media.

### Given Data:
- Refractive index of the lens (μ) = 1.5
- Radius of curvature of the first surface (R1) = -20 cm (concave surface)
- Radius of curvature of the second surface (R2) = +40 cm (convex surface)

### Step 1: Calculate Focal Length in Air
Using the lens maker's formula:
\[ \frac{1}{f} = \left(\mu - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Substituting the values:
\[ \frac{1}{f} = (1.5 - 1) \left(\frac{1}{-20} - \frac{1}{40}\right) \]
\[ \frac{1}{f} = 0.5 \left(-\frac{1}{20} - \frac{1}{40}\right) \]

Finding a common denominator:
\[ -\frac{1}{20} = -\frac{2}{40} \]
Thus,
\[ \frac{1}{f} = 0.5 \left(-\frac{2}{40} - \frac{1}{40}\right) = 0.5 \left(-\frac{3}{40}\right) \]
\[ \frac{1}{f} = -\frac{3}{80} \]

Taking the reciprocal:
\[ f = -\frac{80}{3} \approx 80 \text{ cm} \]

### Step 2: Calculate Focal Length in Liquid (μ = 1.2)
Using the same lens maker's formula:
\[ \frac{1}{f_l} = \left(\mu_l - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Substituting the values:
\[ \frac{1}{f_l} = (1.2 - 1) \left(\frac{1}{-20} - \frac{1}{40}\right) \]
\[ \frac{1}{f_l} = 0.2 \left(-\frac{3}{40}\right) \]
\[ \frac{1}{f_l} = -\frac{0.6}{40} = -\frac{3}{200} \]

Taking the reciprocal:
\[ f_l = -\frac{200}{3} \approx 160 \text{ cm} \]

### Step 3: Calculate Focal Length in Liquid (μ = 2)
Using the same lens maker's formula:
\[ \frac{1}{f_l} = \left(\mu_l - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

Substituting the values:
\[ \frac{1}{f_l} = (2 - 1) \left(\frac{1}{-20} - \frac{1}{40}\right) \]
\[ \frac{1}{f_l} = 1 \left(-\frac{3}{40}\right) \]
\[ \frac{1}{f_l} = -\frac{3}{40} \]

Taking the reciprocal:
\[ f_l = -\frac{40}{3} \approx -160 \text{ cm} \]

### Final Answers:
1. Focal length in air: \( 80 \text{ cm} \)
2. Focal length in liquid (μ = 1.2): \( 160 \text{ cm} \)
3. Focal length in liquid (μ = 2): \( -160 \text{ cm} \)

### Summary of Results:
- Focal length in air: \( 80 \text{ cm} \)
- Focal length in liquid (μ = 1.2): \( 160 \text{ cm} \)
- Focal length in liquid (μ = 2): \( -160 \text{ cm} \)