The dispersive powers of two lenses are `0.01` and `0.02` . If focal length of first lens is `+10cm` , then what should be the focal length of the second lens, so that they form an achromatic combination ?

Options

A. Diverging lens having focal length 20cm
B. Converging lens having focal length 20cm
C. Diverging lens having focal length10cm
D. Converging lens having focal length 10cm

Answer

To find the focal length of the second lens so that the combination of the two lenses forms an achromatic system, we can follow these steps:

### Step 1: Understand the condition for achromatic combination
The condition for two lenses to form an achromatic combination 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 first and second lenses, and \(f_1\) and \(f_2\) are their respective focal lengths.

### Step 2: Substitute the known values
From the problem, we know:
- \(\omega_1 = 0.01\)
- \(f_1 = +10 \, \text{cm}\)
- \(\omega_2 = 0.02\)
- \(f_2 = ?\)

Substituting these values into the achromatic condition gives:
\[ \frac{0.01}{10} + \frac{0.02}{f_2} = 0 \]

### Step 3: Simplify the equation
Calculating \(\frac{0.01}{10}\):
\[ \frac{0.01}{10} = 0.001 \]
Thus, the equation becomes:
\[ 0.001 + \frac{0.02}{f_2} = 0 \]

### Step 4: Rearrange the equation
Rearranging the equation to isolate \(\frac{0.02}{f_2}\):
\[ \frac{0.02}{f_2} = -0.001 \]

### Step 5: Solve for \(f_2\)
Cross-multiplying gives:
\[ 0.02 = -0.001 f_2 \]
Now, solving for \(f_2\):
\[ f_2 = \frac{0.02}{-0.001} = -20 \, \text{cm} \]

### Conclusion
The focal length of the second lens should be \(-20 \, \text{cm}\), indicating that it is a concave lens.