Chapter 18.2, Problem 17PP

To determine

The image and object positions.

Answer to Problem 17PP

The object is placed at a distance of $50\text{mm}$ to the left of the lens and the image is formed at a distance of $50\text{mm}$ to the right of the lens.

Explanation of Solution

Given:

The focal length of the convex lens, $f=25\text{mm}$

The height of the object, $h_o={\text{ h}}_i$

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:

It is given that the image is of the same height as that of the object. This case is possible only when the image formed is real, thus

  $h_i=-h_o$

Negative sign denotes inverted and real image.

Then using the above values in magnification formula, the image and object distance will be,

  $\begin{array}{c}\frac{h_i}{h_o}=-\frac{d_i}{d_o}\\ \frac{-h_o}{h_o}=-\frac{d_i}{d_o}\\ d_i=d_o\mathrm{..........}\text{(1)}\end{array}$

Now, using the given and above values in Thin lens equation, the object distance will be,

  $\begin{array}{c}\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}\\ \frac{1}{25}=\frac{1}{d_o}+\frac{1}{d_o}\\ d_o=2\left(25\right)\\ =50.0\text{mm}\end{array}$

And, then the image distance will be,

  $d_i=d_o=50.0\text{mm}$

Conclusion:

Hence, the object is placed at a distance of $50\text{mm}$ to the left of the lens while the image is formed at a distance of $50\text{mm}$ to the right of the lens.