Question

1 A CONCAVE LENS HAVING RADIUS OF CURVATURE 60CM.
AN OBJECT OF 2M IS PLACED BEFORE 50CM BEFORE CONCAVE
MIRROR. CALCULATE POSITION, SIZE AND NATURE OF IMAGE.
ALSO DRAW RAY DIAGRAM.​

Answer 1

Answer:

Position of image: beyond center of curvature

Size of image: -300 cm

Nature of image: Real Inverted and enlarged

Explanation:

f (focus) = radius of curvature / 2 = 60/2 = 30cm

u = 50cm

height of object = 2m = 200cm

on using mirror formula

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

we get v 75 cm

$magnification(m) = - \frac{v}{u}$

$m = - \frac{75}{50}$

$m \: = - 1.5$

Now,

$m \: = - \frac{v}{u} = \frac{height \: of \: image}{height \: of \: object}$

$- 1.5 = \frac{hight \: of \: image}{200}$

$hight \: of \: image = \: - 300cm$