Chapter 24.6, Problem 1TH

Reproduced below is a side view diagram of the situation described in section II of the tutorial.

Determine an expression for the lateral magnification, $m_1=\raisebox{1ex}{$h^1$}\!\left/ \!\raisebox{-1ex}{$h$}\right.$ , in terms of the object distance, $x_0$ , and the image distance, $x_i$ . (Hint: Draw the principal rays for the tip of the pencil and look for similar triangles. Clearly indicate the similar triangles that you use to determine your answer.)

To determine

To calculate:.

The expression for the lateral magnification $m_l=h^{\prime }/h$ .

Answer to Problem 1TH

The expression for the lateral magnification is $m_l=\frac{h^{\prime }}{h}=\frac{-x_i}{x_o}$.

Explanation of Solution

Given:

An object is placed before a thin lens as follows :

  

Formula used :

In similar triangles ABC and DEF , corresponding sides are in equal proportions or,

AB / BC = DE / EF

AB / AC = DE /DF

Calculation:

Ray diagram showing the principal ray from the tip of pencil that is object, is given below;

  

In the above diagram, two triangles given that are $\Delta OAC$ and $\Delta IGC$ .

are similar.

So that,

  $\frac{IG}{IC}=\frac{OA}{OC}$

Here,

  $IG=h^{\prime }$ ….image height

  $OA=h$ .....object height

And,

  $OC=-x_o$ ......object distance

  $IC=x_i$ ......image distance

Therefore,

  $\begin{array}{l}\frac{h^{\prime }}{x_i}=\frac{h}{-x_o}\\ \frac{h^{\prime }}{h}=\frac{-x_i}{x_0}\end{array}$

Hence, lateral magnification is given below;

  $m_l=\frac{h^{\prime }}{h}=\frac{-x_i}{x_o}$

Conclusion:

Therefore, the lateral magnification, $m_l=\frac{h^{\prime }}{h}=\frac{-x_i}{x_o}$ .