Answer to Question #190269 in Physics for shin

Question

Determine the image distance and image height for a 5.00-cm tall object placed 30.0 cm from a concave mirror having a focal length of 15.0 cm.

1 / f = 1 / do + 1 / di where f =15 cm and do = 45 cm

hi / ho = - di / do where ho = 5 cm, do = 45 cm, and di = 30.0 cm

Answer 1

The lens equation is:

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

where $d_i$ is the unknown image distance. Expressing $d_i$, obtain:

$$\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o} = \dfrac{d_o - f}{fd_o}\\ d_i = \dfrac{fd_o}{d_o - f} = \dfrac{15cm\cdot45cm}{45cm -15cm} = 22.5cm$$

Sign "$-$" means that the image is virtual.

The image height can be found from:

$$\dfrac{h_i}{h_o} = -\dfrac{d_i}{d_o}$$

where $h_i$ is the image height. Expressing $h_i$, find:

$$h_i = -h_o\dfrac{d_i}{d_o} = -\dfrac{5cm\cdot 30cm}{45cm} \approx -3.3cm$$

Answer. Image distance: 22.5cm. Image height: -3.3 cm.