Question

an object 4 cm high is placed at a distance of 6 CM in front of a concave mirror of focal length 12 cm find the position of the image​

Answer 1

Given :

• Object Height (H') = 4 cm
• Object Distance (u) = - 6cm
• Focal length (f) = - 12 cm
• Concave mirror

To FinD :

• Nature, Position of Image

SolutioN :

Use mirror formula for calculating image distance :

⇒1/f = 1/v + 1/u

⇒1/v = 1/f - 1/u

⇒1/v = -1/12 - (-1/6)

⇒1/v = -1/12 + 1/6

⇒1/v = (-1 + 2)/12

⇒1/v = 1/12

⇒v = 12

$\therefore$ Image is formed at a distance of 12cm from mirror.

______________________________

Now, use formula for magnification :

⇒m = -v/u

⇒m = -12/-6

⇒m = 12/6

⇒m = 2

$\therefore$ Magnification of Image is + 2

• Nature of Image is Virtual and erect

Answer 2

$\sf{\underline{Answer\::}}$

The distance from the point of incidence of the mirror to where the image is formed is 12 cm.

$\sf{\underline{Explanation\::}}$

An object kept in front of a concave mirror has a height of 4 cm. The distance between the object and the pole (u) is 6 cm. And the focal length of the concave mirror is -12 cm.

We should find the following,

• Position of the image (v)

We have the mirror formula to be implemented in questions related.

Mirror formula :

• ¹/f = ¹/ᵥ + ¹/ᵤ

Insert the values,

=> $\sf{\dfrac{1}{-12}\:=\:\dfrac{1}{v}\:+\dfrac{1}{-6}}$

=> $\sf{\dfrac{1}{-12}\:=\:\dfrac{-6+v}{-6v}}$

=> $\sf{-6v\:=\:-12(-6+v)}$

=> $\sf{-6v=72-12v}$

=> $\sf{-6v+12v=72}$

=> $\sf{6v=72}$

=> $\sf{v\:=\dfrac{72}{6}}$

=> $\sf{\underline{v=12}}$