Question

An object of height 1.2 is placed before a concave mirror of focal length 20 cm so that a real image is formed at a distance of 60 cm of it . Find the position of a object . What cwill be the height of the image formed

Answer 1

Height of object =h=1.2 cm
Height of image =h'
Focal length=f= -20 cm
Image distance=v= -60 cm ( as image formed is real,it must have formed on same side of object)
Now,
1/v +1/u = 1/f
》1/u = 1/f -1/v = 1/(-20) - 1/(-60)
》1/u = -1/20 + 1/60 = -1/30
》u = -30 cm
Magnification, m = h'/h = -v/u
》h'/1.2 = -(-60)/ (-30)
》h'/1.2 = -2
》h' = -2.4 cm
The negative sign indicates that image formed is inverted.

Answer 2

According to the Question

ho = 1.2 cm

f = - 20 cm

v = - 60 cm

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

Substitute the value of "f" and  "v"

$\bf\huge\frac{1}{u} = \frac{1}{-20} - \frac{1}{-60}$

$\bf\huge\frac{1}{u} = \frac{1}{60} - \frac{1}{20}$

$\bf\huge\frac{1}{u} = \frac{1 - 3}{60}$

$\bf\huge\frac{1}{u} = \frac{-2}{60}$

u = - 30

$\bf\huge\frac{h1}{ho} = \frac{-v}{u}$

$\bf\huge h1 = - \frac{-60}{-30}\times 1.2$

h1 = -2.4 cm