## Question

###### Location of image

What is the location of the final image produced by the compound lens system? Give the x coordinate of the image.

## Answers

Given

Focal lenths f

_{1}= +10.0 cm, f_{2}= +8.00 cmPositions of two lenses x = -20 cm and +20 cm

First we calculate the magnitude only ..temporarily regardlessthe cordinate x

d=50-20=30 cm

distance d' of image S1 produced by len 1(from the len 1)

d'=d.f

_{1}/(d-f_{1})=15 cmS1 is an object to Len 2 with d

_{1}=40-15=25 cm (40 cm is the distance between two lens)

distance of image S2 produced by len 2 (from the len 2)

d''=d

_{1}.f_{2}/(d_{1}-f_{2})= 200/17 cmso the location of final image produced by the system is:200/17 +20 =31.76 cm on x-cordinate

The distance of the object from the first lens is,

$$ p_{1}=50 \mathrm{~cm}-20 \mathrm{~cm}=30 \mathrm{~cm} $$

The thin lens equation is,

$$ \frac{1}{f}=\frac{1}{q}-\frac{1}{p} $$

Rearrange the equation for \(q\)

$$ \begin{aligned} \frac{1}{q} &=\frac{1}{f}+\frac{1}{p} \\ q &=\frac{f p}{f+p} \\ &=\frac{(10.0 \mathrm{~cm})(-30.0 \mathrm{~cm})}{10.0 \mathrm{~cm}-30.0 \mathrm{~cm}} \\ &=15.0 \mathrm{~cm} \end{aligned} $$

The image forms at a distance \(15 \mathrm{~cm}\) right of the first lens. This image acts as object for the second lens.

The separation between the two lenses is,

$$ d=20.0 \mathrm{~cm}+20.0 \mathrm{~cm}=40.0 \mathrm{~cm} $$

The object distance for the second lens is, \(p^{\prime}=40.0 \mathrm{~cm}-15.0 \mathrm{~cm}\)

$$ =25 \mathrm{~cm} $$

The distance of the final image from the second lens can be calculated as follows:

$$ \begin{aligned} q^{\prime} &=\frac{f^{\prime} p^{\prime}}{f^{\prime}+p^{\prime}} \\ &=\frac{(8.00 \mathrm{~cm})(-25 \mathrm{~cm})}{8.00 \mathrm{~cm}+(-25 \mathrm{~cm})} \\ &=-11.76 \mathrm{~cm} \end{aligned} $$

The location of the final image is,

$$ \begin{aligned} x^{\prime} &=20.0 \mathrm{~cm}+11.76 \mathrm{~cm} \\ &=31.76 \mathrm{~cm}=31.8 \mathrm{~cm} \end{aligned} $$