Question

an object of height 1.2 m is placed before a concave mirror of focal length 20cm so that a real image is formed distance 60 cm from it find the position of an object what will be the height of the image formed

Answer 1

Given,
height of object= - 1.2m(By sign convention)
focal length= -20cm(By sign convention)
Image distance= -60cm(By sign convention)
To find,
1)Position of object=?
2)Height of the image formed=?
Solution,
Now by using mirror formula,
1/f = 1/v +1/u
1/u= 1/f-1/v
1/u= (1/-20) - (1/-60)
1/u= 3-1/-60
1/u=2/-60
u=-30
Now by Using Mirror formula. i. e- m= height of image / height of object = v/u
height of image/-1.2= - 60/-30
height of image = - 1.2*2m
= 2.4m
Hope you understand my answer and brainliest it

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