Question

an object of height 20 cm is kept at a distance of 48 CM in front of a mirror of focal length 12cm with the mirror forms a virtual diminished image of object then calculate the distance of the image from the mirror is magnification​

Answer 1

hey mate

here goes your answer

ho =20 cm

f= -12 cm

u =-48 cm

v = ?

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{ - 12} - \frac{1}{ - 48} = \frac{1}{v} \\ \frac{ 3 - 1}{ - 48} = \frac{1}{v} \\ \frac{ - 48}{2} = - 24 = v$

$m = \frac{ - v}{u} \\ = \frac{24}{ - 48} = - 0.5$

hope it helps u ☺☺

Answer 2

The distance of the image from the mirror is 16 cm .

Magnification =1/3

Explanation:

Given : Distance of the object from mirror : u = 48 cm

According to sign- convention.

Focal length = -12 cm  [Mirror is convex mirror as |u|>|f|]

To find : Image distance V

Mirror formula : $\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

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

$\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-48}$

$\dfrac{1}{v}=\dfrac{1}{-12}+\dfrac{1}{48}$

$\dfrac{1}{v}=\dfrac{-4+1}{48}$

$\dfrac{1}{v}=\dfrac{-3}{48}=\dfrac{-1}{16}\Rightarrow\ v= -16$

Hence, the distance of the image from the mirror is 16 cm .

Also, Magnification =$\dfrac{-v}{u}=\dfrac{-(-16)}{48}=\dfrac{1}{3}$

∴ Magnification =1/3

# Learn more :

lease can anyone help me with these questions:

1) An object is 20 cm from a concave mirror of focal length 40 cm. Calculate the distance from the image to the mirror.

2) An object is 40 cm from a concave mirror and the image is formed 60 cm from the mirror. Calculate the magnification.

3) An object is 40 cm from a convex mirror and the image is formed 20 cm from the mirror. Calculate the focal length of the mirror.

4) An object is placed 30 cm in front of a concave mirror. A real image of the object is formed 50 cm from the mirror. What is the focal length of the mirror? if the object is 5 cm high what is the height of the image?

https://brainly.in/question/2624700