Question

An object of 8 cm height is 15 cm in front of a converging lens with focal length of magnitude 4 cm. What is the dimension of the image? Do not include the sign)?

Answer 1

image height is 2.9 cm

Step-by-step explanation:

We have given the size of the object O = 8 cm

Distance of the object from the lens = 15 cm

Focal length of the lens = 4 cm

Now for lens we know that
`(1)/(f)=(1)/(v)-(1)/(u)`

So
`(1)/(4)=(1)/(v)-(1)/(15)`

v = 5.454 cm

Now magnification
`m=(v)/(u)=(5.454)/(15)=0.3636`

Now magnification is the ratio of image height and object height

So
`m=(I)/(O)` , where I is the height of image and O is the height of object

So
`0.3636=(I)/(8)`

I =2.9 cm