The dispersive power of crown and flint glasses are `0.02` and `0.04`, respectively. An achromatic converging lens of focal length `40 (~cm)` is made by keeping two lenses, one of crown glass and the other of flint glass, in contact with each other. The magnitude of product of focal lengths (in cm) of the two lenses are

Answer

To solve the problem, we need to find the product of the focal lengths of the two lenses made of crown glass and flint glass that together form an achromatic lens with a given focal length of 40 cm.

### Step-by-Step Solution:

1. **Understanding the Achromatic Condition**:
For two lenses to be achromatic, the following condition must hold:
\[ \frac{F_1}{\omega_1} + \frac{F_2}{\omega_2} = 0 \]
where \( F_1 \) and \( F_2 \) are the focal lengths of the crown and flint glass lenses, and \( \omega_1 \) and \( \omega_2 \) are their dispersive powers.

2. **Substituting the Values**:
Given that the dispersive power of crown glass (\( \omega_1 \)) is 0.02 and that of flint glass (\( \omega_2 \)) is 0.04, we can substitute these values into the equation:
\[ \frac{F_1}{0.02} + \frac{F_2}{0.04} = 0 \]

3. **Rearranging the Equation**:
Rearranging the equation gives:
\[ \frac{F_1}{0.02} = -\frac{F_2}{0.04} \]
Cross-multiplying yields:
\[ 0.04 F_1 + 0.02 F_2 = 0 \]

4. **Expressing \( F_2 \) in terms of \( F_1 \)**:
From the equation, we can express \( F_2 \):
\[ F_2 = -2 F_1 \]

5. **Using the Combined Focal Length**:
The combined focal length \( F \) of the two lenses is given by:
\[ \frac{1}{F} = \frac{1}{F_1} + \frac{1}{F_2} \]
Substituting \( F_2 = -2 F_1 \) into the equation gives:
\[ \frac{1}{40} = \frac{1}{F_1} + \frac{1}{-2 F_1} \]

6. **Finding a Common Denominator**:
The equation can be simplified:
\[ \frac{1}{40} = \frac{1 - 0.5}{F_1} = \frac{0.5}{F_1} \]

7. **Solving for \( F_1 \)**:
Rearranging gives:
\[ F_1 = 0.5 \times 40 = 20 \text{ cm} \]

8. **Finding \( F_2 \)**:
Now substituting back to find \( F_2 \):
\[ F_2 = -2 \times 20 = -40 \text{ cm} \]

9. **Calculating the Product of Focal Lengths**:
The magnitude of the product of the focal lengths is:
\[ |F_1 \times F_2| = |20 \times -40| = 800 \text{ cm}^2 \]

### Final Answer:
The magnitude of the product of the focal lengths of the two lenses is \( 800 \text{ cm}^2 \).

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