Two identical thin planoconvex glass lenses (refractive index `1.5)` each having radius of curvature of `20cm` are placed with their convex surfaces 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`s
B. `-25cm`
C. `50cm`
D. `50cm`

Answer

To solve the problem, we need to find the focal length of the combination of two identical plano-convex lenses and the intervening oil layer. We will use the lens maker's formula for each lens and then combine their focal lengths.

### Step 1: Identify the parameters
- Refractive index of glass lenses, \( \mu_g = 1.5 \)
- Refractive index of oil, \( \mu_o = 1.7 \)
- Radius of curvature for both lenses, \( R = 20 \, \text{cm} \)

### Step 2: Calculate the focal length of the first plano-convex lens (Lens 1)
Using the lens maker's formula:
\[ \frac{1}{f_1} = (\mu_g - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a plano-convex lens:
- \( R_1 = +20 \, \text{cm} \) (convex side)
- \( R_2 = \infty \) (plane side)

Substituting the values:
\[ \frac{1}{f_1} = (1.5 - 1) \left( \frac{1}{20} - 0 \right) = 0.5 \times \frac{1}{20} = \frac{1}{40} \]
Thus,
\[ f_1 = 40 \, \text{cm} \]

### Step 3: Calculate the focal length of the second plano-convex lens (Lens 2)
Since the lenses are identical:
\[ f_2 = f_1 = 40 \, \text{cm} \]

### Step 4: Calculate the focal length of the oil layer (acting as a concave lens)
Using the lens maker's formula again for the oil layer:
\[ \frac{1}{f_3} = (\mu_o - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the oil layer:
- \( R_1 = -20 \, \text{cm} \) (convex side of the first lens)
- \( R_2 = +20 \, \text{cm} \) (convex side of the second lens)

Substituting the values:
\[ \frac{1}{f_3} = (1.7 - 1) \left( \frac{1}{-20} - \frac{1}{20} \right) = 0.7 \left( -\frac{1}{20} - \frac{1}{20} \right) = 0.7 \left( -\frac{2}{20} \right) = -\frac{0.7}{10} = -0.07 \]
Thus,
\[ f_3 = -\frac{10}{0.7} \approx -14.29 \, \text{cm} \]

### Step 5: Combine the focal lengths
Using the formula for the combination of lenses:
\[ \frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3} \]
Substituting the values:
\[ \frac{1}{f_{\text{eq}}} = \frac{1}{40} + \frac{1}{40} - \frac{0.07}{1} \]
Finding a common denominator (which is 40):
\[ \frac{1}{f_{\text{eq}}} = \frac{1 + 1 - 2.8}{40} = \frac{2 - 2.8}{40} = \frac{-0.8}{40} \]
Thus,
\[ f_{\text{eq}} = -\frac{40}{0.8} = -50 \, \text{cm} \]

### Final Answer
The focal length of the combination is \( -50 \, \text{cm} \).

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