The dispersive powers of glasses of lenses used in an achromatic pair are in the ratio 5 : 3. If the focal length of the concave lens is 15 cm , then the nature and focal length of the other lens would be

Options

A. convex, 9 cm
B. concave, 9 cm
C. convex, 25 cm
D. concave, 25 cm

Explanation

To solve the problem step by step, we will follow the concepts of dispersive power and the formula for an achromatic lens pair.

### Step 1: Understand the Given Information
We have two lenses in an achromatic pair. The dispersive powers of the lenses are in the ratio of 5:3. We denote the dispersive power of the first lens (concave lens) as \( \omega_1 = 5x \) and the second lens (unknown) as \( \omega_2 = 3x \). The focal length of the concave lens is given as \( f_1 = -15 \, \text{cm} \).

### Step 2: Use the Formula for Achromatic Pair
The formula for an achromatic pair of lenses is given by:
\[ \frac{\omega_1}{f_1} + \frac{\omega_2}{f_2} = 0 \]
Substituting the values we have:
\[ \frac{5x}{-15} + \frac{3x}{f_2} = 0 \]

### Step 3: Simplify the Equation
Rearranging the equation gives:
\[ \frac{3x}{f_2} = \frac{5x}{15} \]
We can simplify \( \frac{5x}{15} \) to \( \frac{x}{3} \):
\[ \frac{3x}{f_2} = \frac{x}{3} \]

### Step 4: Solve for \( f_2 \)
Now, we can cross-multiply to solve for \( f_2 \):
\[ 3x \cdot 3 = x \cdot f_2 \]
This simplifies to:
\[ 9x = x \cdot f_2 \]
Assuming \( x \neq 0 \), we can divide both sides by \( x \):
\[ f_2 = 9 \, \text{cm} \]

### Step 5: Determine the Nature of the Lens
Since \( f_2 \) is positive, the lens is a convex lens.

### Final Answer
The nature of the other lens is convex, and its focal length is \( 9 \, \text{cm} \).