A concavo-convex lens has refractive index `1.5` and the radii of curvature of its surfaces are `10 cm and 20 cm.` The concave surface is upwards and is filled with oil of refractive index `1.6.` The focal length of the combination will be

Options

A. `18.18 cm`
B. `15 cm`
C. `22 cm`
D. `28.57 cm`

Explanation

To find the focal length of the concavo-convex lens filled with oil, we will use the lens maker's formula. Here is the step-by-step solution:

### Step 1: Understand the Lens Configuration
We have a concavo-convex lens with:
- Refractive index of the lens, \( \mu = 1.5 \)
- Radii of curvature: \( R_1 = 10 \, \text{cm} \) (concave surface, upwards) and \( R_2 = -20 \, \text{cm} \) (convex surface, downwards)
- Refractive index of oil, \( \mu_{\text{oil}} = 1.6 \)

### Step 2: Apply the Lens Maker's Formula
The lens maker's formula is given by:

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

### Step 3: Calculate the Focal Length for the Oil-Filled Surface
For the oil-filled concave surface, we will first calculate the focal length \( f_1 \):

\[ \frac{1}{f_1} = (\mu_{\text{oil}} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the values:

\[ \frac{1}{f_1} = (1.6 - 1) \left( \frac{1}{10} - \frac{1}{-20} \right) \]

Calculating the terms:

\[ \frac{1}{f_1} = 0.6 \left( \frac{1}{10} + \frac{1}{20} \right) \]

Finding a common denominator:

\[ \frac{1}{10} + \frac{1}{20} = \frac{2}{20} + \frac{1}{20} = \frac{3}{20} \]

Thus,

\[ \frac{1}{f_1} = 0.6 \cdot \frac{3}{20} = \frac{1.8}{20} = \frac{0.09}{1} \]

So,

\[ f_1 = \frac{1}{0.09} \approx 11.11 \, \text{cm} \]

### Step 4: Calculate the Focal Length for the Lens
Now, we will calculate the focal length \( f_2 \) for the lens itself:

\[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the values:

\[ \frac{1}{f_2} = (1.5 - 1) \left( \frac{1}{10} - \frac{1}{-20} \right) \]

Calculating the terms:

\[ \frac{1}{f_2} = 0.5 \left( \frac{1}{10} + \frac{1}{20} \right) \]

Finding a common denominator:

\[ \frac{1}{10} + \frac{1}{20} = \frac{3}{20} \]

Thus,

\[ \frac{1}{f_2} = 0.5 \cdot \frac{3}{20} = \frac{1.5}{20} = \frac{0.075}{1} \]

So,

\[ f_2 = \frac{1}{0.075} \approx 13.33 \, \text{cm} \]

### Step 5: Combine the Focal Lengths
Now, we combine the two focal lengths using the formula for two thin lenses in contact:

\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \]

Substituting the values:

\[ \frac{1}{f} = \frac{1}{11.11} + \frac{1}{13.33} \]

Calculating:

\[ \frac{1}{f} \approx 0.09 + 0.075 = 0.165 \]

Thus,

\[ f \approx \frac{1}{0.165} \approx 6.06 \, \text{cm} \]

### Final Answer
The focal length of the combination is approximately \( 6.06 \, \text{cm} \).

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