Question

An object of height is 4cm is kept at distance 12cm in front of a mirror it's virtual and erect image size 1.5cm is obtain find the focal length and nature of the mirror also draw the ray diagram ​

Answer 1

Answer:-

Size of object( h2) = 4cm

Size of image (h1) = 1.5cm

Step by step Explaination:-

Given:-

Object distance(u) = 12cm

Image distance(v) = ?????

Formula-:

Magnification (m) = (-v/u) = (h2/h1)

=> (-v/u) = (1. 5/4)

=> (-v/12) = (15/40)

=> (-v) = (15*12/40)

=> (-v) = 3*12/8

=> - v = 4.5cm

=> v = -4.5cm

As mirror formula,

=> 1/v + 1/u = 1/f

=> -1/4.5 + 1/12 = 1/f

=> -12 + 4.5/4.5*12 = 1/f

=> -7.5/54 = 1/f

=> f = (-54/7.5)

=> f = (-540/75)

=> f = -7.2cm

Verification-:

As the focal length is negative so it is a convex mirror that is forming a virtual and erect image.

Answer 2

Explanation:

Solution is in the attachment