Two equi convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in & contact and space between them is filled with water of refractive index `4/3`. The combination works as

Options

A. converging lens of focal length 30 cm
B. diverging lens of focal length 15 cm
C. converging lens of focal length 15 cm
D. diverging lens of focal length 40 cm

Explanation

To solve the problem, we need to determine the effective focal length of the combination of two equi-convex lenses placed in contact, with water filling the space between them. Here’s a step-by-step solution:

### Step 1: Identify the Parameters
- Focal length of each lens (f₁ and f₂): 20 cm
- Refractive index of the lenses (μ₁): 1.5
- Refractive index of water (μ₂): 4/3

### Step 2: Calculate the Radius of Curvature
For equi-convex lenses, the radius of curvature (R) can be calculated using the lens maker's formula:

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

For equi-convex lenses, \(R_1 = R\) and \(R_2 = -R\). Therefore, we have:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = 2(\mu - 1) \frac{1}{R} \]

Rearranging gives:

\[ R = 2(\mu - 1)f \]

Substituting the values:

\[ R = 2(1.5 - 1)(20) = 2(0.5)(20) = 20 \text{ cm} \]

### Step 3: Calculate the Effective Focal Length of the Combination
When two lenses are in contact, the effective focal length (F) can be calculated using the formula:

\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{(μ_2 - 1)}{R} \]

Substituting the values:

- \(f_1 = f_2 = 20 \text{ cm}\)
- \(μ_2 = \frac{4}{3}\)
- \(R = 20 \text{ cm}\)

Calculating:

\[ \frac{1}{F} = \frac{1}{20} + \frac{1}{20} - \frac{\left(\frac{4}{3} - 1\right)}{20} \]

Calculating \(μ_2 - 1\):

\[ \frac{4}{3} - 1 = \frac{1}{3} \]

Now substituting back:

\[ \frac{1}{F} = \frac{1}{20} + \frac{1}{20} - \frac{\frac{1}{3}}{20} \]

Calculating:

\[ \frac{1}{F} = \frac{1}{20} + \frac{1}{20} - \frac{1}{60} \]

Finding a common denominator (60):

\[ \frac{1}{F} = \frac{3}{60} + \frac{3}{60} - \frac{1}{60} = \frac{5}{60} = \frac{1}{12} \]

Thus, the effective focal length \(F\) is:

\[ F = 12 \text{ cm} \]

### Conclusion
The combination of the two equi-convex lenses with water in between acts as a lens with an effective focal length of 12 cm.

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