Question

A convex mirror with a radius of curvature of 0.450 m is placed above theaisles in a store. Determine the image distance and magnification of acardboard box on the floor 2.75 m below the mirror. Is the image virtual orreal? Is the image inverted or upright?

Answer 1

Given:

The radius of the curvature of the convex mirror is

`R=0.450\text{ m}`

The distance of the object is

`u=-2.75\text{ m}`

Required: image distance and the magnification

Explanation:

first, we find the focal length of the mirror by the relation is

`f=(R)/(2)`

plugging the values in the above, we get

`\begin{gathered} f=\frac{0.450\text{ m}}{2} \\ f=0.22\text{5 m} \end{gathered}`

the mirror formula is given by

`(1)/(f)=(1)/(v)+(1)/(u)`

Plugging all the values in the above relation, we get

`\begin{gathered} \frac{1}{0.225\text{ m}}=(1)/(v)+\frac{1}{-2.75\text{ m}} \\ (1)/(v)=(1)/(0.225)+(1)/(2.75) \\ (1)/(v)=(2.75+0.225)/(2.75*0.225) \\ (1)/(v)=(2.975)/(0.618) \\ v=0.21\text{ m} \end{gathered}`

the distance of the image is 0.21 m. it will be formed behind the mirror.

now calculate the magnification. magnification is given by

`m=-(v)/(u)`

Plugging all the values in the above, we get

`\begin{gathered} m=-\frac{0.21m}{-2.75\text{ m}} \\ m=0.076\text{ } \end{gathered}`

The magnification is 0.076.

the image formed by the convex mirror is virtual, inverted, and erect.