Answer to Question #116926 in Optics for Eromobor Emmanuella

Question

An object 0.04m high is placed in front of and at right angle to the axis of a concave mirror which has radius of curvature 0.15m. if a real image 0.12m high is obtained determine by scale drawing or otherwise the distance of the object from the mirror. How far and in what direction must the object be moved in order to obtain a virtual image 0.12m high.

Answer 1

1)

$$f=0.5(0.15)=0.075\ m$$

In our case, we can write formula for concave mirror as

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\to \frac{1}{0.075}=\frac{1}{0.12}+\frac{1}{u}$$

$$u=0.2\ m$$

2) In our case, we have $f=0.075 m, u=-0.12 m$

$$\frac{1}{0.075}=-\frac{1}{0.12}+\frac{1}{v}$$

$$v=0.046\ m$$

The object must be placed between the mirror and its focus.