Chapter 34: Q. 35 (page 991)

A $2.0 - c m$ -tall object is $15 c m$ in front of a converging lens that has a $20 c m$ focal length.
$(a)$ Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram.
$(b)$ Calculate the image position and height. Compare with your ray-tracing answers in part $a$ .

Short Answer

Expert verified

Part $(a)$

$(a)$ The image distance is $s^{'} = - 60 c m$ and image height is $h^{'} = 8.0 c m .$

Part $(b)$

$(b)$ The image distance is $s^{'} = - 60 c m$ and image height is $h^{'} = 8.0 c m$

Step by step solution

Step: 1 Finding lateral magnification:

From thin lens equation,

$\frac{1}{s} + \frac{1}{s^{'}} = \frac{1}{f}$

The convex lens of focal lens is positive and concave lens is negative:the image virutal is negative and real image distance is positive.

The lateral magnification is

$| M | = \frac{image height}{object height} = \frac{h^{'}}{h} M = - \frac{s^{'}}{s}$

object height is $h = 2.0 c m .$

The lens is converging.

The distance lens object is $s = 15 c m .$

The focal length is $f = 20 c m .$

Step: 2 Finding image distance: (part a)

The ray trace diagram is

From lens equation is

$\begin{array}{l} \frac{1}{s} + \frac{1}{s^{'}} = \frac{1}{f} \\ \frac{1}{s^{'}} = \frac{1}{f} - \frac{1}{s} \\ \frac{1}{s^{'}} = \frac{s - f}{s f} \\ s^{'} = \frac{s f}{s - f} \\ s^{'} = \frac{(15 cm) \times (20 cm)}{15 cm - 20 cm} \\ s^{'} = - 60 cm . \end{array}$

Step: 3 Finding image height: (part b)

The lateral magnification lens by

$\begin{array}{l} M = - \frac{s^{'}}{s} \\ M = - \frac{- 60 cm}{15 cm} \\ M = 4 . \\ \begin{array}{l} | M | = \frac{h^{'}}{h} \\ h^{'} = h | M | \\ h^{'} = (2.0 cm) \times (4) \\ h^{'} = 8.0 cm . \end{array} \end{array}$