Question

A convex mirror with a focal length of 0.25 m forms a 0.080 m tall image of an automobile at a distance of 0.24 m behind the mirror. What is the magnification of the image? Where is the car located, and what is its height? Is the image real or virtual? Is the image upright or inverted? Draw a ray diagram to show where the image forms and how large it is with respect to the object

Answer 1

Distance of the object = 6 m

Height of the object = 2 m

Step-by-step explanation:

Thinking process:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Therefore, using formula of lens :

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

`(1)/(u) = (1)/(6)`

solving, gives u = 6

The magnification is calculated as follows:

m = -0.24/-6

= 0.04

The height = 2 m

The diagram yields an image behind the mirror which is upright.

Answer 2

The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.

Step-by-step explanation:

Given that,

Focal length = 0.25 m

Length of image = 0.080 m

Image distance = 0.24 m

We need to calculate the distance of the object

Using formula of lens

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

Put the value into the formula

`(1)/(0.24)=(1)/(0.25)+(1)/(u)`

`(1)/(u)=(1)/(0.24)-(1)/(0.25)`

`(1)/(u)=(1)/(6)`

`u=6\ m`

We need to calculate the magnification

Using formula of magnification

`m=-(v)/(u)`

Put the value into the formula

`m=-(0.24)/(-6)`

`m=0.04`

We need to calculate the height of the object

Using formula of magnification

`m=(h')/(h)`

`h=(0.080)/(0.04)`

`h=2\ m`

A convex mirror produce a virtual and upright image behind the mirror.

Hence, The distance and height of the object is 6 m and 2 m.

The image is virtual and upright.