Question

an object 7m high is placed 15cm from a convex spherical mirror of radius of curvature 45cm.describe its image

Answer 1

Let's first find the image distance of the convex mirror from the mirror equation:

$$\dfrac{1}{d_o}+\dfrac{1}{d_i}=-\dfrac{1}{f},$$ $$\dfrac{1}{d_o}+\dfrac{1}{d_i}=-\dfrac{2}{R},$$ $$d_i=\dfrac{1}{-\dfrac{2}{R}-\dfrac{1}{d_o}},$$ $$d_i=\dfrac{1}{-\dfrac{2}{45\ cm}-\dfrac{1}{15\ cm}}=-9\ cm=-0.09\ m.$$

The sign minus means that the image is virtual and formed behind the mirror.

Let's find the image height from the magnification equation:

$$\dfrac{h_i}{h_o}=-\dfrac{d_i}{d_o},$$ $$h_i=h_o\dfrac{-d_i}{d_o}=7\ m\cdot\dfrac{-(-0.09\ m)}{0.15\ m}=4.2\ m.$$

The sign plus means that the image is upright and diminished.

Answer:

The image is 4.2 m high, virtual, upright, diminished and formed behind the mirror.