Question

(ii) An object at a distance of 16 cm from a spherical mirror forms a virtual image at a distance of 12 cm behind the mirror. Determine the magnification of the image and type of the mirror.

Hint:

For virtual images formed by mirrors, the image distance is positive for concave mirrors and negative for convex mirrors. The magnification is negative for real images and positive for virtual images.

Answer 1

We are given:
- Object distance \( u = -16 \, \text{cm} \) (since the object is in front of the mirror)
- Image distance \( v = +12 \, \text{cm} \) (since the image is virtual and behind the mirror, the distance is positive)
We can use the mirror formula to calculate the focal length: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substituting the given values: \[ \frac{1}{f} = \frac{1}{12} - \frac{1}{-16} = \frac{1}{12} + \frac{1}{16} \] Finding the common denominator: \[ \frac{1}{f} = \frac{4 + 3}{48} = \frac{7}{48} \] Thus, the focal length is: \[ f = \frac{48}{7} \approx 6.86 \, \text{cm} \] Since the focal length is positive, the mirror is a concave mirror (a converging mirror).
Next, the magnification \( m \) can be calculated using the magnification formula: \[ m = \frac{v}{u} = \frac{12}{-16} = -0.75 \] So, the magnification is \( -0.75 \), meaning the image is diminished and virtual.