A diverging meniscus lens of radii of curvatures 25 cm and 50 cm has a refractive index 1.5. Its focal length is (in cm)

Options

A. `-50`
B. `-100`
C. `100`
D. `50`

Answer

To find the focal length of a diverging meniscus lens with given radii of curvature and refractive index, we can use the lens maker's formula:

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of curvature \( r_1 = -25 \, \text{cm} \) (negative because it's a diverging lens)
- Radius of curvature \( r_2 = -50 \, \text{cm} \) (also negative for a diverging lens)
- Refractive index \( \mu = 1.5 \)

2. **Write the lens maker's formula:**
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

3. **Substitute the values into the formula:**
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{-25} - \frac{1}{-50} \right) \]

4. **Calculate \( \mu - 1 \):**
\[ \mu - 1 = 0.5 \]

5. **Calculate \( \frac{1}{r_1} - \frac{1}{r_2} \):**
\[ \frac{1}{-25} - \frac{1}{-50} = -\frac{1}{25} + \frac{1}{50} \]
To combine these fractions, find a common denominator (which is 50):
\[ -\frac{2}{50} + \frac{1}{50} = -\frac{1}{50} \]

6. **Substitute back into the formula:**
\[ \frac{1}{f} = 0.5 \left( -\frac{1}{50} \right) = -\frac{0.5}{50} = -\frac{1}{100} \]

7. **Find the focal length \( f \):**
\[ f = -100 \, \text{cm} \]

### Conclusion:
The focal length of the diverging meniscus lens is \( f = -100 \, \text{cm} \).

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