Question

A 3.10-cm diameter coin is placed a distance of 25.0cm from a concave lens that has a focal length of -11.0 cm. Determine the image distance ad diameter of the image.Must follow GRESA format. Make sure to list the L-O-S-T Image formed accordingly Solution format:Given Required: Equation(s) Needed: Solution:(a) (b) Final Answer L-ocation: O-rientation: S-ize: T-ype:

Answer 1

Given

3.1 cm diameter coin

25 cm from the concave lens

11 cm focal lens

Procedure

Below is a sketch how the image is created using a concave lens:

Equation(s) Needed:

$\begin{gathered} \frac{1}{o}+\frac{1}{i}=\frac{1}{f} \\ \frac{1}{i}=\frac{1}{f}-\frac{1}{o} \end{gathered}$

Solution

$\begin{gathered} \frac{1}{i}=\frac{1}{-11}-\frac{1}{-25} \\ i=-7.64\text{ cm} \end{gathered}$

The negative sign means that the image is virtual.

(b)

$\begin{gathered} \frac{h}{i}=\frac{H}{o} \\ h=\frac{H}{o}\cdot i \\ h=3.1\text{ cm}\cdot\frac{7.64\text{ cm}}{25\text{ cm}} \\ h=0.95\text{ cm } \end{gathered}$

0.95 cm - the diameter of the coin’s image

Final answer:

L-ocation: -7.64 cm

O-rientation: Inverted

S-ize: 0.95 cm

T-ype: Virtual