Question

A 1.5-cm object is placed 0.50 m to the left of a diverging lens with a focal length of 0.20 m. A converging lens with a focal length of 0.17 m is located 0.08 m to the right of the diverging lens. What is the height and orientation with respect to the original object of the final image?

Answer 1

Answer:

The object for the converging lens is upright and 0.429 cm tall, the image of this converging lens is inverted and 1.375 cm high

Explanation:

Let

$d_{o}=distance of object[tex]\\f=focal length\\d_{i}=distance of image\\I_{h}=Image height$

For diverging lens:

$d_{o} = 0.50\\f = -0.20\\\frac{1}{d_{o}}+\frac{1}{-0.20}\\\frac{1}{d_{i}}=\frac{1}{-0.20}-\frac{1}{0.50}=-7\\d_{o}=-\frac{1}{7}$

Magnification = $\frac{d_{i}}{d_{o}}= -\frac{1}{7}÷ 0.5 = -0.286$

Image height $= -0.286 * 1.5 = -0.429 cm$ (negative sign means the image is virtual, inverted.

This image is $\frac{1}{7}$ meter to left of the center of the diverging lens.

The converging lens is located 0.08 m to the right of the diverging lens

The distance between the image of the diverging lens and center of the converging lens = $\frac{1}{7} + 0.08 = 0.229 m$

The image of the diverging lens becomes the object of the converging lens.

$d_{o} = 0.223\\f = 0.17\\\frac{1}{d_{i}}=\frac{1}{0.17}-\frac{1}{0.223}=0.715\\d_{i}=0.715m to the right of the converging lens$

$Magnification =\frac{d_{i}}{d_{o}} = \frac{0.715}{0.223}=3.206\\image height=3.206 * 0.429 = 1.375 cm.$