Question

Determine the image distance and image height for a 8.00-cm tall object placed 46.5 cm from a convex lens having a focal length of 16.0 cm. 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) (c) Final Answer L-ocation: O-rientation: S-ize:T-ype:

Answer 1

Given:

• Height of image = 8.00 cm

,

• Distance from convex = 46.5 cm

,

• Focal length = 16.0 cm

Let's find the image distance.

To find the image distance, apply the formula below.

Equation:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Required:

f = 16.0 cm

u = 46.5 cm

Let's solve for v

We have:

$\begin{gathered} \frac{1}{16.0}=\frac{1}{46.5}+\frac{1}{v} \\ \\ \frac{1}{v}=\frac{1}{16.0}-\frac{1}{46.5} \\ \\ \frac{1}{v}=\frac{46.5-16}{744}=\frac{30.5}{744} \\ \\ v=\frac{744}{30.5} \\ \\ v=24.4\text{ cm} \end{gathered}$

Now apply the formula to find the height:

$\begin{gathered} h^{\prime}=\frac{-v\times h}{u} \\ \\ h^{\prime}=\frac{-24.4\times8.00}{46.5} \\ \\ h^{\prime}=-4.2\operatorname{cm} \end{gathered}$

Therefore, the distance is 24.4 cm

The height of the object is -4.2 cm

ANSWER:

Given: f = 16.0 cm; u = 46.5 cm, h = 8.00 cm

Required: v and h'

Equations needed:

$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ \\ h^{\prime}=\frac{-v\times h}{u} \end{gathered}$

Solution: v = 24.4 cm

h' = -4.2 cm

L-ocation: Behind the mirror

O-rientation: Inverted

S-ize: Reduced

T-ype: Real