Question

solve ...............​

Answer 1

refer to the attachment

Answer 2

ANSWER:-

Given:

A 5cm tall object is placed at a distance of 30cm from a convex mirror of focal length 15cm.

To find:

Find the position, size & nature of the image formed.

Solution:

⏺️h1= +5

⏺️u= -30cm

⏺️f= +15cm

⏺️v= ?

Using the formula of the mirror;

$= > \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ = > \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ = > \frac{1}{v} = \frac{1}{15} - \frac{1}{( - 30)} \\ \\ = > \frac{1}{v} = \frac{1}{15} + \frac{1}{30} \\ \\ = > \frac{1}{v} = \frac{2 + 1}{30} \\ \\ = > \frac{1}{v} = \frac{3}{30} \\ \\ = > \frac{1}{v} = \frac{1}{10} \\ \\ = > v = + 10$

Now,

⏺️The ratio of the height of the image to the height of the object is Known as linear magnification.

$m = \frac{h2 }{h1} = \frac{ - v}{u} \\ \\ = > \frac{h2}{5} = \frac{ - ( + 10)}{ - 30} \\ \\ = > h2 = \frac{50}{30} \\ \\ = > h2 = + 1.66cm$

⏺️Nature= Virtual, erect.

Hope it helps ☺️