Question

A candle is placed 18 cm in front of a concave mirror to generate a real, inverted and doubly magnified image. The radius of curvature of the concave mirror is ................ cm. (answer in integer)

Hint:

The magnification for concave mirrors is given by \( m = -\frac{v}{u} \). Use the mirror equation to find the focal length and then use \( R = 2f \) to find the radius of curvature.

Answer 1

Step 1: Understanding the relationship for magnification. For a concave mirror, the magnification \( m \) is given by: \[ m = \frac{-v}{u} \] Where:
- \( v \) is the image distance
- \( u \) is the object distance
- Given that the image is real, inverted, and doubly magnified, the magnification \( m = -2 \). So, we have: \[ -2 = \frac{-v}{18} \quad \Rightarrow \quad v = 36 \, \text{cm} \] Step 2: Using the mirror equation. The mirror equation relates the object distance \( u \), image distance \( v \), and the focal length \( f \): \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Since \( v = 36 \) cm and \( u = -18 \) cm (object is in front of the mirror), we substitute these values into the equation: \[ \frac{1}{f} = \frac{1}{36} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{1}{36} - \frac{1}{18} = \frac{-1}{36} \] \[ f = -36 \, \text{cm} \] Step 3: Relating focal length to radius of curvature. The radius of curvature \( R \) is related to the focal length \( f \) by: \[ R = 2f \] So, the radius of curvature is: \[ R = 2 \times (-36) = -72 \, \text{cm} \] Final Answer: \[ \boxed{-72 \, \text{cm}} \]