Identify the given values:
- Refractive index of the lens material, $\mu = 1.5$.
- Radii of curvatures, $R_1 = 12 \, \text{cm}$ and $R_2 = 24 \, \text{cm}$.
Determine the signs of the radii of curvature using sign convention:
- We assume light travels from left to right.
- According to the problem, the "convex face receiving incident rays" means the first surface (where light enters) is convex.
- For a convex surface on the left, its center of curvature lies to its right. Therefore, the radius of curvature $R_1$ is positive.
  
  $$ R_1 = +12 \, \text{cm} $$
- The lens is a "concavo-convex lens", which is typically also called a meniscus lens. In a meniscus lens, both centers of curvature lie on the same side of the lens. Since the center of curvature for the first (convex) surface is to the right, the center of curvature for the second (concave) surface must also be to the right.
- For a concave surface with its center of curvature to the right, the radius of curvature $R_2$ is also positive.
  
  $$ R_2 = +24 \, \text{cm} $$
Apply the Lens Maker's Formula:

The focal length $f$ of a thin lens in air is given by:

$$ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$
Substitute the values into the formula:

$$ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{+12 \, \text{cm}} - \frac{1}{+24 \, \text{cm}} \right) $$

$$ \frac{1}{f} = (0.5) \left( \frac{1}{12 \, \text{cm}} - \frac{1}{24 \, \text{cm}} \right) $$
Perform the calculation:

To subtract the fractions, find a common denominator, which is 24.

$$ \frac{1}{f} = (0.5) \left( \frac{2}{24 \, \text{cm}} - \frac{1}{24 \, \text{cm}} \right) $$

$$ \frac{1}{f} = (0.5) \left( \frac{2 - 1}{24 \, \text{cm}} \right) $$

$$ \frac{1}{f} = (0.5) \left( \frac{1}{24 \, \text{cm}} \right) $$

$$ \frac{1}{f} = \frac{0.5}{24 \, \text{cm}} $$

$$ \frac{1}{f} = \frac{1/2}{24 \, \text{cm}} $$

$$ \frac{1}{f} = \frac{1}{2 \times 24 \, \text{cm}} $$

$$ \frac{1}{f} = \frac{1}{48 \, \text{cm}} $$
Calculate the focal length:

$$ f = 48 \, \text{cm} $$

Therefore, the focal length of the lens is 48 cm.

The final answer is $\boxed{\text{48 cm}}$.

Ray Optics

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