Lenses are constructed by a material of refractive indeic 1'50. The magnitude of the radii of curvature are 20 cm and 30 cm. Find the focal lengths of the possible lenses with the above specifications.

Explanation

To solve the problem of finding the focal lengths of the possible lenses constructed from a material with a refractive index of 1.5 and given radii of curvature of 20 cm and 30 cm, we will use the lens maker's formula. The lens maker's formula is given by:

\[ \frac{1}{f} = \left( \frac{\mu_l - 1}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( \mu_l \) is the refractive index of the lens material,
- \( \mu_m \) is the refractive index of the medium (1 for air),
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 1: Identify the lens types and their configurations
We will consider four possible lens configurations:
1. **Double Convex Lens (Convex-Convex)**
2. **Concave-Convex Lens (Convex-Concave)**
3. **Double Concave Lens (Concave-Concave)**
4. **Concave-Convex Lens (Concave-Convex)**

### Step 2: Calculate the focal length for each lens type

#### Case 1: Double Convex Lens
- For a double convex lens, \( R_1 = +20 \, \text{cm} \) and \( R_2 = -30 \, \text{cm} \).
- Using the lens maker's formula:
\[ \frac{1}{f} = \left( \frac{1.5 - 1}{1} \right) \left( \frac{1}{20} - \frac{1}{-30} \right) \]
\[ = 0.5 \left( \frac{1}{20} + \frac{1}{30} \right) \]
Calculating \( \frac{1}{20} + \frac{1}{30} \):
\[ = \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \]
Thus,
\[ \frac{1}{f} = 0.5 \cdot \frac{1}{12} = \frac{1}{24} \]
So, \( f = 24 \, \text{cm} \).

#### Case 2: Concave-Convex Lens
- For a concave-convex lens, \( R_1 = +20 \, \text{cm} \) and \( R_2 = +30 \, \text{cm} \).
- Using the lens maker's formula:
\[ \frac{1}{f} = \left( \frac{1.5 - 1}{1} \right) \left( \frac{1}{20} - \frac{1}{30} \right) \]
Calculating \( \frac{1}{20} - \frac{1}{30} \):
\[ = \frac{3 - 2}{60} = \frac{1}{60} \]
Thus,
\[ \frac{1}{f} = 0.5 \cdot \frac{1}{60} = \frac{1}{120} \]
So, \( f = 120 \, \text{cm} \).

#### Case 3: Double Concave Lens
- For a double concave lens, \( R_1 = -20 \, \text{cm} \) and \( R_2 = +30 \, \text{cm} \).
- Using the lens maker's formula:
\[ \frac{1}{f} = \left( \frac{1.5 - 1}{1} \right) \left( \frac{-1}{20} - \frac{1}{30} \right) \]
Calculating \( -\frac{1}{20} - \frac{1}{30} \):
\[ = -\left( \frac{3 + 2}{60} \right) = -\frac{5}{60} = -\frac{1}{12} \]
Thus,
\[ \frac{1}{f} = 0.5 \cdot -\frac{1}{12} = -\frac{1}{24} \]
So, \( f = -24 \, \text{cm} \).

#### Case 4: Concave-Concave Lens
- For a concave-concave lens, \( R_1 = -20 \, \text{cm} \) and \( R_2 = -30 \, \text{cm} \).
- Using the lens maker's formula:
\[ \frac{1}{f} = \left( \frac{1.5 - 1}{1} \right) \left( \frac{-1}{20} - \frac{-1}{30} \right) \]
Calculating \( -\frac{1}{20} + \frac{1}{30} \):
\[ = -\frac{3 - 2}{60} = -\frac{1}{60} \]
Thus,
\[ \frac{1}{f} = 0.5 \cdot -\frac{1}{60} = -\frac{1}{120} \]
So, \( f = -120 \, \text{cm} \).

### Summary of Focal Lengths
1. **Double Convex Lens:** \( f = +24 \, \text{cm} \)
2. **Concave-Convex Lens:** \( f = +120 \, \text{cm} \)
3. **Double Concave Lens:** \( f = -24 \, \text{cm} \)
4. **Concave-Concave Lens:** \( f = -120 \, \text{cm} \)