Two identical thin planoconvex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex sufaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7 The focal length of the combination is

Options

A. `-20cm`
B. `-25cm`
C. `-50cm`
D. 50cm

Explanation

To find the focal length of the combination of two identical plano-convex lenses with oil in between, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters:**
- Refractive index of glass lenses, \( \mu_g = 1.5 \)
- Refractive index of oil, \( \mu_o = 1.7 \)
- Radius of curvature of the lenses, \( R = 20 \, \text{cm} \)

2. **Calculate the Focal Length of Each Glass Lens:**
- For a plano-convex lens, the focal length \( f \) can be calculated using the formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) \]
- Since the plano side is flat, the formula simplifies to:
\[ \frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{20} \right) \]
- Thus:
\[ \frac{1}{f_1} = 0.5 \times \frac{1}{20} = \frac{0.5}{20} = \frac{1}{40} \]
- Therefore, the focal length of each glass lens is:
\[ f_1 = 40 \, \text{cm} \]

3. **Calculate the Focal Length of the Oil Layer:**
- The focal length for the oil layer between the two lenses can be calculated similarly:
\[ \frac{1}{f_2} = (\mu_o - 1) \left( \frac{1}{-R} + \frac{1}{R} \right) \]
- This simplifies to:
\[ \frac{1}{f_2} = (1.7 - 1) \left( \frac{1}{-20} + \frac{1}{20} \right) \]
- This results in:
\[ \frac{1}{f_2} = 0.7 \left( \frac{-1 + 1}{20} \right) = 0.7 \left( \frac{-2}{20} \right) = -\frac{0.7}{10} = -\frac{7}{100} \]
- Therefore, the focal length of the oil layer is:
\[ f_2 = -\frac{100}{7} \approx -14.29 \, \text{cm} \]

4. **Combine the Focal Lengths:**
- The total focal length of the combination of the two lenses and the oil layer can be calculated using the formula:
\[ \frac{1}{f_{\text{equiv}}} = \frac{1}{f_1} + \frac{1}{f_2} \]
- Substituting the values:
\[ \frac{1}{f_{\text{equiv}}} = \frac{1}{40} + \left(-\frac{7}{100}\right) \]
- Finding a common denominator (200):
\[ \frac{1}{f_{\text{equiv}}} = \frac{5}{200} - \frac{14}{200} = -\frac{9}{200} \]
- Therefore, the effective focal length is:
\[ f_{\text{equiv}} = -\frac{200}{9} \approx -22.22 \, \text{cm} \]

### Final Answer:
The focal length of the combination is approximately \( -22.22 \, \text{cm} \).