Question

A grasshopper hops to a point on the central axis of a spherical mirror. The absolute magnitude of the mirror’s focal length is $\mathbf{40}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, and the lateral magnification of the image produced by the mirror is $\mathbf{+}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{200}$. (a) Is the mirror convex or concave? (b) How far from the mirror is the grasshopper?

Short answer

1. It is a convex mirror.
2. Object distance, $p=160cm$

Step-by-step solution

Determine the given quantities

The magnitude of focal length, $f=40.0cm$

Lateral magnification, $m=+0.2$

Determine the concepts of magnification

Using the formula for lateral magnification, we can find out the sign of image distance. If it is negative, then one can guess the sign of the focal length. Using the mirror formula, object distance can be calculated.

Formula:

Mirror’s formula$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

Lateral magnification$m=-\frac{i}{p}$

(a) Calculate the type of the lens

Consider the lateral magnification as

$\begin{array}{c}m=-\frac{i}{p}\\ =+0.2\end{array}$

Therefore, $i=-0.2\times p$, $i<0,$and the image is diminished in size, which implies that the spherical mirror has $f=-40.0cm.$

Therefore, it is a convex mirror.

(b) Calculate the object distance

Consider the mirror’s formula:

$\frac{1}{p}+\frac{1}{i}=\frac{1}{f}$

Substitute$iandf$in the above equation and solve as:

$\frac{1}{p}-\frac{10}{2p}=-\frac{1}{40}$

$\frac{2}{2p}-\frac{10}{2p}=-\frac{1}{40}$

$\frac{2-10}{2p}=-\frac{1}{40}$

$p=160cm$

Object distance,$p=160cm$