Two lenses in contact made of materials with dispersive powers in the ratio`2:1`behaves as ab achromatic lens of focal length `10cm`.The individual focal length of the lenses are :

Options

A. 5 cm, -10cm
B. `-5cm , 10cm`
C. 10 cm, -20 cm
D. `-20 cm, 10 cm`

Explanation

To solve the problem of finding the individual focal lengths of two lenses in contact that behave as an achromatic lens, we can follow these steps:

### Step 1: Understand the Condition for Achromatism
The condition for achromatism for two lenses in contact is given by the formula:
\[ \frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0 \]
where \(\omega_1\) and \(\omega_2\) are the dispersive powers of the two lenses, and \(f_1\) and \(f_2\) are their respective focal lengths.

### Step 2: Use the Given Ratio of Dispersive Powers
We are given that the dispersive powers are in the ratio \(2:1\). Thus, we can express this as:
\[ \omega_1 = 2k \quad \text{and} \quad \omega_2 = k \]
for some constant \(k\).

### Step 3: Substitute into the Achromatism Condition
Substituting the expressions for \(\omega_1\) and \(\omega_2\) into the achromatism condition gives:
\[ \frac{2k}{f_1} + \frac{k}{f_2} = 0 \]
This simplifies to:
\[ 2 \cdot \frac{1}{f_1} + \frac{1}{f_2} = 0 \]
From this, we can express \(f_1\) in terms of \(f_2\):
\[ \frac{1}{f_2} = -\frac{2}{f_1} \implies f_1 = -2f_2 \]

### Step 4: Use the Focal Length of the Achromatic Lens
The net focal length \(F\) of two lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
Given that \(F = 10 \, \text{cm}\), we have:
\[ \frac{1}{10} = \frac{1}{f_1} + \frac{1}{f_2} \]

### Step 5: Substitute \(f_1\) into the Focal Length Equation
Substituting \(f_1 = -2f_2\) into the focal length equation gives:
\[ \frac{1}{10} = \frac{1}{-2f_2} + \frac{1}{f_2} \]
This simplifies to:
\[ \frac{1}{10} = -\frac{1}{2f_2} + \frac{1}{f_2} \]
Combining the fractions on the right side:
\[ \frac{1}{10} = \frac{-1 + 2}{2f_2} = \frac{1}{2f_2} \]

### Step 6: Solve for \(f_2\)
Cross-multiplying gives:
\[ 2f_2 = 10 \implies f_2 = 5 \, \text{cm} \]

### Step 7: Find \(f_1\)
Using \(f_1 = -2f_2\):
\[ f_1 = -2 \times 5 = -10 \, \text{cm} \]

### Conclusion
The individual focal lengths of the lenses are:
- \(f_1 = -10 \, \text{cm}\)
- \(f_2 = 5 \, \text{cm}\)