Question

An object is placed in front of a diverging lens, such that the object-to-image distance is 71 cm.

Find the focal length of the lens, if the image is 26 cm from the lens. The image is real.

A virtual object lies 95 cm in front of a diverging lens, with a 29 cm focal length.
Find the location of the image.
Find the magnification
Is the image real or virtual?

Answer 1

Step-by-step explanation:

Given that,

Object-to-image distance d= 71 cm

Image distance = 26 cm

We need to calculate the object distance

`u -v= d`

`u=71+26=97\ cm`

We need to calculate the focal length

Using formula of lens

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

put the value into the formula

`(1)/(f)=(1)/(-26)+(1)/(97)`

`(1)/(f)=-(71)/(2522)`

`f=-35.52\ cm`

The focal length of the lens is 35.52.

(B). Given that,

Object distance = 95 cm

Focal length = 29 cm

We need to calculate the distance of the image

Using formula of lens

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

Put the value in to the formula

`(1)/(-29)=(1)/(v)-(1)/(95)`

`(1)/(v)=(1)/(-29)-(1)/(95)`

`(1)/(v)=-(124)/(2755)`

`v=-22.21\ cm`

We need to calculate the magnification

Using formula of magnification

`m=(v)/(u)`

`m=(22.21)/(95)`

`m=0.233`

The magnification is 0.233.

The image is virtual.

Hence, This is the required solution.