A diverging lens is made of material with refractive index `1.3` and has identical concaves surfaces of radius `20 cm.` The lens is immersed in a transparent medium with refractive index `1.8.`
(a) What is now the focal length of the lens?
(b) What is the minimum distance taht an immersed object must be from the lens so that a real image is formed?

Explanation

To solve the problem step by step, we will break it down into two parts as specified in the question.

### Part (a): Finding the Focal Length of the Lens

1. **Understand the Lensmaker's Formula**:
The lensmaker's formula is given by:
\[ \frac{1}{f} = \frac{\mu_1}{\mu_2 - 1} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( f \) = focal length of the lens
- \( \mu_1 \) = refractive index of the lens material
- \( \mu_2 \) = refractive index of the medium
- \( R_1 \) = radius of curvature of the first surface
- \( R_2 \) = radius of curvature of the second surface

2. **Identify Given Values**:
From the problem:
- \( \mu_1 = 1.3 \) (refractive index of the lens)
- \( \mu_2 = 1.8 \) (refractive index of the medium)
- \( R_1 = -20 \) cm (since it's a concave surface, we take it as negative)
- \( R_2 = 20 \) cm (the second surface is also concave)

3. **Substitute Values into the Formula**:
Substitute the values into the lensmaker's formula:
\[ \frac{1}{f} = \frac{1.3}{1.8 - 1} \left( \frac{1}{-20} - \frac{1}{20} \right) \]

4. **Calculate the Right Side**:
- Calculate \( \mu_2 - 1 = 1.8 - 1 = 0.8 \)
- Calculate \( \frac{1}{-20} - \frac{1}{20} = -\frac{1}{20} - \frac{1}{20} = -\frac{2}{20} = -\frac{1}{10} \)

Now substitute these values:
\[ \frac{1}{f} = \frac{1.3}{0.8} \left( -\frac{1}{10} \right) \]

5. **Simplify**:
\[ \frac{1}{f} = \frac{1.3 \times -1}{0.8 \times 10} = \frac{-1.3}{8} = -0.1625 \]

6. **Find the Focal Length**:
\[ f = \frac{1}{-0.1625} \approx -6.15 \text{ cm} \]
Since the focal length is negative, it confirms that it is a diverging lens.

### Part (b): Finding the Minimum Distance for a Real Image

1. **Understand the Condition for Real Image Formation**:
For a diverging lens, a real image is formed when the object is placed at a distance greater than the focal length from the lens.

2. **Identify the Focal Length**:
From part (a), we found that the focal length \( f \approx -6.15 \) cm.

3. **Determine the Minimum Object Distance**:
For a real image to be formed, the object distance \( u \) must satisfy:
\[ |u| > |f| \]
Since \( f \) is negative, we take the absolute value:
\[ |u| > 6.15 \text{ cm} \]

4. **Conclusion**:
Therefore, the minimum distance that an immersed object must be from the lens to form a real image is:
\[ u > 6.15 \text{ cm} \]

### Final Answers:
(a) The focal length of the lens is approximately **-6.15 cm**.
(b) The minimum distance that an immersed object must be from the lens to form a real image is **greater than 6.15 cm**.