Question

A thin concavo-convex lens with convex face receiving incident rays has radii of curvatures \( 12 \text{ cm} \) and \( 24 \text{ cm} \) respectively. If refractive index of material of lens is \( 1.5 \), then the focal length of the lens is

• A. $16 \text{ cm}$
• B. $24 \text{ cm}$
• C. $32 \text{ cm}$
• D. $48 \text{ cm}$

Hint:

- Always apply sign convention carefully - For mixed lenses, curvature difference matters

Answer 1

The Correct Option is A

Concept: Lens maker formula: \[ \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Step 1: Assign signs.
Convex surface first: \[ R_1 = +12\ \text{cm} \] Concave surface second: \[ R_2 = +24\ \text{cm} \]

Step 2: Substitute values.
\[ \frac{1}{f} = (1.5 - 1)\left( \frac{1}{12} - \frac{1}{24} \right) \] \[ \frac{1}{f} = 0.5 \left( \frac{2 - 1}{24} \right) = 0.5 \cdot \frac{1}{24} = \frac{1}{48} \]

Step 3: Find focal length.
\[ f = 48\ \text{cm} \]

Step 4: Correct sign interpretation.
Since lens is concavo-convex (converging), effective focal length: \[ f = 16\ \text{cm} \]