The dispersive powers of two materials are `0.30 & 0.28`. They are used to construct two lenses which are kept in contact to eliminate chromatic aberration (that means the `f_(v)=f_(r)` the focal ) If the material of disperive power 0.30 is 10cm, then the focal lenght (for av, color) of the lens of other material is :

Options

A. 28/3 cm
B. `-0.20//3m`
C. 0.75/7m
D. None of these

Explanation

To solve the problem, we need to find the focal length of the second lens (f2) given that the lenses are in contact and eliminate chromatic aberration. The dispersive powers of the materials are given as 0.30 and 0.28, and the focal length of the first lens (f1) is 10 cm.

### Step-by-Step Solution:

1. **Understanding the Concept**:
When two lenses are in contact and designed to eliminate chromatic aberration, the condition that must be satisfied is:
\[ \frac{w_1}{f_1} + \frac{w_2}{f_2} = 0 \]
where \(w_1\) and \(w_2\) are the dispersive powers of the two materials, and \(f_1\) and \(f_2\) are the focal lengths of the respective lenses.

2. **Assigning Values**:
From the problem, we know:
- \(w_1 = 0.30\) (dispersive power of the first lens)
- \(f_1 = 10 \, \text{cm}\) (focal length of the first lens)
- \(w_2 = 0.28\) (dispersive power of the second lens)
- \(f_2\) is what we need to find.

3. **Substituting Values into the Equation**:
Substitute the known values into the equation:
\[ \frac{0.30}{10} + \frac{0.28}{f_2} = 0 \]

4. **Solving for \(f_2\)**:
Rearranging the equation gives:
\[ \frac{0.28}{f_2} = -\frac{0.30}{10} \]
\[ \frac{0.28}{f_2} = -0.03 \]
Now, cross-multiply to solve for \(f_2\):
\[ 0.28 = -0.03 f_2 \]
\[ f_2 = \frac{0.28}{-0.03} \]
\[ f_2 = -\frac{28}{3} \, \text{cm} \]

5. **Final Result**:
The focal length of the second lens is:
\[ f_2 = -\frac{28}{3} \, \text{cm} \approx -9.33 \, \text{cm} \]

### Conclusion:
The focal length of the lens made from the material with a dispersive power of 0.28 is approximately \(-9.33 \, \text{cm}\).