A convex-concave diverging lens is made of glass of refractive index 1.5 and focal length 24 cm. radius of curvature for one surface is double that of the other. Then radii of curvature for the two surfaces are (in cm )

Options

A. 6,12
B. 12,24
C. 3,6
D. 18,36

Answer

To find the radii of curvature for the two surfaces of a convex-concave diverging lens, we can follow these steps:

### Step 1: Understand the lens formula
The lens formula for a thin lens is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( f \) is the focal length of the lens,
- \( n \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 2: Assign values and signs
Given:
- The focal length \( f = -24 \) cm (negative for diverging lens),
- The refractive index \( n = 1.5 \),
- The relationship between the radii of curvature \( R_1 = 2R_2 \).

### Step 3: Substitute values into the lens formula
Substituting the known values into the lens formula:
\[ \frac{1}{-24} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{-24} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 4: Substitute \( R_1 \) in terms of \( R_2 \)
Since \( R_1 = 2R_2 \), we can substitute this into the equation:
\[ \frac{1}{-24} = 0.5 \left( \frac{1}{2R_2} - \frac{1}{R_2} \right) \]
This becomes:
\[ \frac{1}{-24} = 0.5 \left( \frac{1 - 2}{2R_2} \right) = 0.5 \left( \frac{-1}{2R_2} \right) = \frac{-1}{4R_2} \]

### Step 5: Solve for \( R_2 \)
Now, we can solve for \( R_2 \):
\[ \frac{1}{-24} = \frac{-1}{4R_2} \]
Cross-multiplying gives:
\[ -4R_2 = -24 \implies R_2 = 6 \text{ cm} \]

### Step 6: Find \( R_1 \)
Using the relationship \( R_1 = 2R_2 \):
\[ R_1 = 2 \times 6 = 12 \text{ cm} \]

### Final Answer
The radii of curvature for the two surfaces are:
- \( R_1 = 12 \) cm
- \( R_2 = 6 \) cm