Question

An object with a height of 0.42 m is placed 0.50 m in front of a mirror with a focal length of 1.00 m.

a) Determine the location of the image.

b) Determine the magnification.

c) Determine the size (height) of the image.

Explain choice by using sign convention.

d) Is the image real or virtual?

e) Is it upright or inverted?

f) Is it enlarged or reduced?

Answer 1

here,

the height og object , ho = 0.42 m

the object distance , do = 0.5 m

the focal length , f = 1 m

a)

let the location of image be di

using the mirror formula

1/f = 1/di + 1/do

1/1 = 1/di + 1/0.5

di = -1 m

the location of image is 1 m behind the mirror

b)

the magnification , m = - di /do

m = - (-1)/(0.5) = 2

c)

the image height , hi = ho * m

hi = 0.42 * 2 m = 0.84 m

d)

as the image distance is negative

the image is virtual

e)

as the image height is positive

the image is upright

f)

as di > do

the image is enlarged