Question

A square wire of side 3 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? The centre of the wire is on the axis of the with it's two sides normal to the axis.​

Answer 1

Answer:

• Area enclosed by the image (-2cm) ² = 4cm²

Step-by-step explanation:

u = -25cm, f = -10cm, v = ?

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$

$= - \frac{1}{10} - \frac{1}{ - 25} \\ = \frac{1}{25} - \frac{1}{10}$

v = -16.67 cm

Magnification,

$m = \frac{1}{o} = - \frac{v}{u}$

$\frac{1}{3} = - \frac{ - 16.67}{ - 25} \: \: or \: \: \\ l = - \frac{16.67 \times 3}{25} cm \: \: or \: \: \\ l \: = - 2cm$

• Area enclosed by the image (-2cm) ² = 4cm²