An object of length 4 cm is placed in front of a concave mirror at a distance 30 cm. The focal length of mirror is 15 cm. (a) Where will the image form? (b) What will be the length of image?

(a) Given,

u = 30 cm (negative)

f = 15 cm (negative)

Mirror formula:

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Substituting the values in the formula above, we get,

$- \dfrac{1}{30} + \dfrac{1}{v} = - \dfrac{1}{15} \\[0.5em] \dfrac{1}{v} = - \dfrac{1}{15} + \dfrac{1}{30} \\[0.5em] \dfrac{1}{v} = \dfrac{ - 2 + 1 }{30} \\[0.5em] \dfrac{1}{v} = -\dfrac{1}{30} \\[0.5em] \Rightarrow v = 30 \text{ cm} \\[0.5em]$

Hence, the image is formed at 30 cm in front of the mirror

(b) Magnification

$\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \dfrac{I}{4} = - \dfrac{30}{30} \\[0.5em] \dfrac{I}{4} = - 1 \\[0.5em] \Rightarrow I = - 4 \text { cm}\\[0.5em]$

Hence, length of image = 4 cm

Negative sign represents that the image is inverted.