Answer to Question #190272 in Physics for shin

Question

 Determine the image distance and image height for a 5.00-cm tall object placed 10.0 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 = 10.0 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:

$$\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o} = \dfrac{d_o - f}{fd_o}\\ d_i = \dfrac{fd_o}{d_o - f} = \dfrac{15cm\cdot 10cm}{10cm -15cm} = -30cm$$

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: -30cm. Image height: 3.3 cm.