Chapter 18.2, Problem 16PP

To determine

The object position and object height.

Answer to Problem 16PP

  $19.6\text{cm}$ to the left of the lens and $3.4\text{cm}$ , respectively.

Explanation of Solution

Given:

The focal length of the convex lens, $f=6.8\text{cm}$ .

The height of the image, $h_i=-1.8\text{cm}$ (since the real image is formed, hence it will be inverted and thereby will be of negative sign).

The image distance from the lens, $d_i=10.4\text{cm}$ (since the real image is formed, hence it will be formed to the right of the lens and thereby will be of positive sign).

Formula used:

Thin lens equation,

  $\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Where,

  $\begin{array}{l}f=\text{focal length of the lens}\\ d_i=\text{ image distance}\\ d_o=\text{object distance}\end{array}$

And,

Magnification, $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$

Where,

  $\begin{array}{l}{h}_i=\text{image height}\\ h_o=\text{object height}\end{array}$

Calculation:

Using the given values in Thin lens equation, the object distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{6.8}=\frac{1}{10.4}+\frac{1}{d_o}\\ d_o=\frac{10.4\times 6.8}{10.4-6.8}\\ =19.6\text{cm}\end{array}$

Then, the object height will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{-1.8}{h_o}=-\frac{10.4}{19.6}\\ h_i=3.4\text{cm}\end{array}$

Conclusion:

Hence, the object position is $19.6\text{cm}$ to the left of the lens and object height is $3.4\text{cm}$ .