Question

Consider the mirror shown here. An object 4 cm tall stands 10 cm in front of a converging mirror of focal length 5 cm.

object
The image formed by this mirror is: (Select all that apply.)
real
virtual
enlarged
the same size
diminished
Inverted
erect

Answer 1

Real

The same size

Inverted

Answer 2

Real

The same size

Inverted

Step-by-step explanation:

Given,

Object's height, h₀ = 4 cm

The distance between the object and mirror, u = 10 cm

The focal length of the mirror, f = 5 cm

The mirror formula is given by the relation

`(1)/(f) =(1)/(u) +(1)/(v)`

Substituting the given values in the above equation according to the sign convention

`(1)/(-5) =(1)/(-10) +(1)/(v)`

v = -10 cm

Therefore, the image distance from the concave mirror is v = 10 cm from the mirror, which is at the same spot of the object.

The magnification factor is given by the formula

M = - v/u = h₁/h₀

Where

h₁/h₀ = - v/u

Therefore the height of the image

h₁ = -(v/u)h₀

Substituting the values in the equation according to the sign convention

h₁ = -(-10/-10)x4

= -4 cm

Therefore, the height of the image is, h₁ = -4 cm

This implies that the image formed is the same size as the object, and the negative indicates the image formed is inverted.