Q. 37 A聽2.0-cm-tall object is聽15聽cm... [FREE SOLUTION]

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Chapter 34: Q. 37 (page 991)

A $2.0 - c m -$ tall object is $15 c m$ in front of a diverging 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

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Part $(a)$

$(a)$ The image distance is $s^{'} = - 8.6 c m$ and image height is $h^{'} = 1.1 c m .$

Part $(b)$

$(b)$ The image distance is $s^{'} = - 8.6 c m$ and image height is $h^{'} = 1.1 c m .$

Both part agrees the condition.

Step by step solution

Step: 1 Finding lateral magnification:

From thin lens equation,

$\frac{1}{s} + \frac{1}{s^{鈥�}} = \frac{1}{f}$

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

$| M | = \frac{image height}{object height} = \frac{h^{鈥�}}{h} M = 鈭� \frac{s^{鈥�}}{s}$

The object height is $h = 2.0 c m .$

The lens is diverging.

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

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

Step: 2 Finding image distance: (part a)

From lens equation is

$\begin{array}{l} \frac{1}{s} + \frac{1}{s^{鈥�}} = \frac{1}{f} \\ \frac{1}{s^{鈥�}} = \frac{1}{f} 鈭� \frac{1}{s} \\ \frac{1}{s^{鈥�}} = \frac{s 鈭� f}{s f} \\ s^{鈥�} = \frac{s f}{s 鈭� f} \\ s^{鈥�} = \frac{(15 cm) 脳 (鈭� 20 cm)}{15 cm 鈭� (鈭� 20 cm)} \\ s^{鈥�} = 鈭� 8.6 cm . \end{array}$

The ray trace diagram is

Step: 3 Finding image height: (part b)

The lateral magnification lens by

$\begin{array}{l} M = 鈭� \frac{s^{鈥�}}{s} \\ M = 鈭� \frac{鈭� 8.6 cm}{15 cm} \\ M = 鈭� \frac{4}{7} . \end{array} \begin{array}{l} | M | = \frac{h^{鈥�}}{h} \\ h^{鈥�} = h | M | \\ h^{鈥�} = (2.0 cm) 脳 (\frac{4}{7}) \\ h^{鈥�} = 1.1 cm . \end{array}$