Question

An object is kept on principal axid of a concave mirror of focal length 12 cm . If the object at a distance of 18 cn from the mirror .Calculate image distance and magnification of the mirror .​

Answer 1

Given :

▪ Focal length = 12cm

▪ Distance of object =18cm

▪ Type of mirror : concave

To Find :

▪ Distance of image

▪ Magnification

Concept :

X-coordinate of centre of curvature and focus of concave mirror are negative and those forconvex mirror are positive. In case of mirror since light rays reflect back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

✴ Mirror formula :

1/u + 1/v = 1/f

✴ Formula of lateral magnification :

m = - v/u

Calculation :

❇ Distance of image :

→ 1/u + 1/v = 1/f

→ 1/(-18) + 1/v = 1/(-12)

→ 1/v = -1/12 + 1/18

→ 1/v = (-3+2)/36

→ 1/v = -1/36

→ v = -36 cm

❇ Lateral magnification :

→ m = - v/u

→ m = - (-36)/(-18)

→ m = -2

Answer 2

Given:-

Type of Mirror:- Concave

Focal length= 12 cm

Object distance= 18 cm

To Find:-

Image distance

Concept:-

X co-ordinates the focus of concave mirror are negative and centre of curvature and also the focus of convex mirror is positive.

Formula used :-

$\large\implies{\tt{\green{\frac{1}{u} + \frac{1}{v}= \frac{1}{f}}}}$

lateral magnification:-

$\large\implies{\tt{m=\frac{-v}{u}}}$

Calculations:-

Distance of the image

$\large\implies{\tt{\red{\frac{1}{u} + \frac{1}{v}= \frac{1}{f}}}}$

$\large\implies{\tt{\red{\frac{1}{-18} + \frac{1}{v}= \frac{1}{-12}}}}$

$\large\implies{\tt{\red{\frac{1}{v}= \frac{1}{-12} +\frac{1}{18}}}}$

$\large\implies{\tt{\red{\frac{1}{v} = \frac{(-3+2)}{36}}}}$

$\large\implies{\tt{\red{\frac{1}{v} = \frac{(-1)}{36}}}}$

$\large\implies{\tt{\purple{v = -36cm }}}$

Magnification of the mirror

$\large\implies{\tt{\pink{m= \frac{-v}{u}}}}$

$\large\implies{\tt{\pink{m= \cancel\frac{- 36}{-18}}}}$

$\large\implies{\tt{\purple{m=2}}}$