Four lenses are made from the same type of glass, the radius of curvature of each face is given below. Which will have the greatest positive power

Options

A. 10 cm convex and 15 cm concave
B. 5 cm convex and 10 cm concave
C. 15 cm convex and plane
D. 20 cm convex and 30 cm concave

Explanation

To determine which of the four lenses has the greatest positive power, we will use the lens maker's formula:

\[ \frac{1}{f} = (μ - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( μ \) is the refractive index of the lens material,
- \( r_1 \) and \( r_2 \) are the radii of curvature of the two lens surfaces.

Since all lenses are made from the same type of glass, \( μ \) will be constant for all lenses. Therefore, we can focus on the term \( \frac{1}{r_1} - \frac{1}{r_2} \) to find the lens with the greatest positive power, as power \( P \) is given by:

\[ P = \frac{1}{f} \]

### Step-by-step Solution:

1. **Identify the radii of curvature for each lens:**
- Lens A: \( r_1 = 10 \, \text{cm} \) (convex), \( r_2 = -15 \, \text{cm} \) (concave)
- Lens B: \( r_1 = 5 \, \text{cm} \) (convex), \( r_2 = -10 \, \text{cm} \) (concave)
- Lens C: \( r_1 = 15 \, \text{cm} \) (convex), \( r_2 = \infty \) (plane)
- Lens D: \( r_1 = 20 \, \text{cm} \) (convex), \( r_2 = -30 \, \text{cm} \) (concave)

2. **Calculate \( \frac{1}{f} \) for each lens using the lens maker's formula:**
- **Lens A:**
\[ \frac{1}{f_A} = \frac{1}{10} - \left(-\frac{1}{15}\right) = \frac{1}{10} + \frac{1}{15} \]
\[ = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \, \text{(positive)} \]

- **Lens B:**
\[ \frac{1}{f_B} = \frac{1}{5} - \left(-\frac{1}{10}\right) = \frac{1}{5} + \frac{1}{10} \]
\[ = \frac{2}{10} + \frac{1}{10} = \frac{3}{10} \, \text{(positive)} \]

- **Lens C:**
\[ \frac{1}{f_C} = \frac{1}{15} - \left(-\frac{1}{\infty}\right) = \frac{1}{15} - 0 = \frac{1}{15} \, \text{(positive)} \]

- **Lens D:**
\[ \frac{1}{f_D} = \frac{1}{20} - \left(-\frac{1}{30}\right) = \frac{1}{20} + \frac{1}{30} \]
\[ = \frac{3}{60} + \frac{2}{60} = \frac{5}{60} = \frac{1}{12} \, \text{(positive)} \]

3. **Compare the values of \( \frac{1}{f} \) for each lens:**
- Lens A: \( \frac{1}{6} \)
- Lens B: \( \frac{3}{10} \)
- Lens C: \( \frac{1}{15} \)
- Lens D: \( \frac{1}{12} \)

4. **Determine which lens has the greatest positive power:**
- Convert to a common denominator to compare:
- Lens A: \( \frac{5}{30} \)
- Lens B: \( \frac{9}{30} \)
- Lens C: \( \frac{2}{30} \)
- Lens D: \( \frac{2.5}{30} \)

The greatest value is for Lens B, which has a power of \( \frac{3}{10} \).

### Conclusion:
The lens with the greatest positive power is **Lens B** with a power of \( \frac{3}{10} \).