Question

An object with a height of −0.060 m points below the principal axis (it is inverted) and is 0.120 m in front of a diverging lens. The focal length of the lens is −0.28 m. (Include the sign of the value in your answers.) What is the image distance? What is the magnification? What is the image height?

Answer 1

Step-by-step explanation:

Given that,

Height of object = -0.060 m

Distance of object = 0.120 m

Focal length = -0.28 m

We need to calculate the image distance

Using formula of lens

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

Where, f = focal length

v = image distance

u = object distance

Put the value into the formula

`(1)/(-0.28)=(1)/(v)+(1)/(-0.120)`

`(1)/(v)=(1)/(-0.28)-(1)/(0.120)`

`(1)/(v)=-(250)/(21)`

`v=-0.084\ m`

Negative sign shows the image on the same side of the object and the image is virtual.

(b). We need to calculate the magnification

Using formula of magnification

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

`m=(0.084)/(0.120)`

`m=0.7`

Positive value of magnification shows the object and image is inverted.

(c). We need to calculate the image height

Using formula of magnification

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

`0.7=(h')/(-0.060)`

`h'=0.7*(-0.060)`

`h'=-0.042\ m`

The height of the image is 0.042 m.

Hence, This is the required solution.