Question

A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens of the same side as the object. (a) What is the focal length of the lens? Is the lens converging of diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect of inverted?

Answer 1

The focal length of the lens and the height of the image are -6.86 cm and -0.638 cm.

Step-by-step explanation:

Given that,

Distance of the object u = -16.0 cm

Distance of the image v = -12.0 cm

Height of the object = 8.50 mm = 0.85 cm

(a). We need to calculate the focal length

Using lens's formula

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

Where, f = focal length

u = distance of the object

v = distance of the image

Put the value into the formula

`(1)/(f)=(1)/(-16.0)+(1)/(-12.0)`

`(1)/(f)=-(7)/(48)`

`f=-(48)/(7)`

`f=-6.86\ cm`

The lens is diverging.

(b). Using formula of magnification

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

`-(v)/(u)=(h')/(h)`

Put the value in to the formula

`-(12.0)/(16.0)=(h')/(0.85)`

`h'=-(12.0*0.85)/(16.0)`

`h'=-0.638\ cm`

The image is inverted.

Hence, The focal length of the lens and the height of the image are -6.86 cm and -0.638 cm.