Question

Two converging lenses with focal lengths of $40cm$and $20cm$are $10cm$apart. A $2.0cm$tall object is $15cm$in front of the$40cm$focal-length lens.

a. Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.

b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

Short answer

a. The statement is given below.

b. The image position and height. Compare with your ray-tracing answers in part a. is $s{\text{'}}_1=-24cm,s{\text{'}}_2=48.57cmandh\text{'}=4.58cm$.

Step-by-step solution

Part (a) step 1: Given Information 

We have to given the focal length of $40cm$and $20cm$are$10cm$.

We need to find ray tracing to find the position and height of the image.

Part (a) step 2: Simplify

Given diagram is two converging lenses. When tracing the rays in these lenses, the rays are still parallel as it passes through the two lenses.

The value are known as:

• $f_1=40cm$
  
• $f_2=20cm$
  
• $d=10cm$
  
• $h=2cm$
  
• $s_1=15cm$
  
• $s_2=24+10=34cm$
  

Part (b) step 1: Given Information

We need to Calculate the image position and height. Compare with your ray-tracing answers in part a.

Part (b) step 2: Simplify

First, solving for the image distance using the formula:

$s\text{'}=\frac{sf}{s-f}$

Now, we will determine the magnification of the image using the formula:

$m=-\frac{s\text{'}}{s}$

We used to determine the height,by the magnification to the formula:

$h\text{'}=m_1{m}_{2}h$

Part (b) step 3:Calculation

Then, solving for the first lens,

$$\begin{gathered} s_1^{\text{'}}==\frac{s_1{f}_1}{s_1-f_1} \\ =\frac{\left(15\right)\left(40\right)}{15-40} \\ =-24\text{cm} \\ {m}_1=-\frac{s_1^{\text{'}}}{s_1}=-\frac{-24}{15} \\ =1.6\text{cm} \end{gathered}$$

Also, solving for the second lens,

$$\begin{gathered} s_2^{\text{'}}==\frac{s_2{f}_2}{s_2-f_2} \\ =\frac{\left(34\right)\left(20\right)}{34-\left(20\right)} \\ =48.57\text{cm} \\ {m}_2=-\frac{s_2^{\text{'}}}{s_2}=-\frac{48.57}{34} \\ =-1.43\text{cm} \end{gathered}$$

Solving for the height of the image

$$\begin{gathered} h\text{'}=m_1{m}_{2}h \\ =\left(1.6\right)\left(-1.43\right)\left(2\right) \\ =-4.58cm \end{gathered}$$

Then, the height is$4.58cm.$