Question

An object of size 2 cm is placed 8 cm to the left of a converging lens with the focal length of 6cm. A diverging lens of focal length 12 cm is placed a distance 18 cm to the right of the converging lens. Describe the final image (location, size, orientation, and type)

Answer 1

Final image is located at 1.71 cm to the left of diverging lens. Final image will be inverted and virtual.

Step-by-step explanation:

In this question,

we have to find location of final image,v=?

In this question we have given,

object distance from converging lens,u=-8cm

focal length of converging lens,
`f_(c)`=6cm

distance between converging and diverging lens=18cm

focal length of diverging(concave) lens,
`f_(d)`=-12cm

To find the location of final image, we will first find the location of image formed by converging(convex) lens

we know that u, v and
`f_(c)` are related by following formula

`(1)/(f_(c)) =(1)/(v)- (1)/(u).`............(1)

put values of
`f_(c)` and u in equation (1)

we got,

`(1)/(6) =(1)/(v)- (1)/(-8)`

`(1)/(6)-(1)/(8) =(1)/(v)`

`(8-6)/(6* 8) =(1)/(v)`

`v=(6* 8)/(2)\\ v=24cm`

The image is located 24 cm to the right of the converging lens. This image is real and inverted.

Now, image formed by converging lens will act as object for the diverging lens which is now located at (18-20=-2)cm at the left hand side of diverging lens

it means,

u=-2cm

we know that u, v and
`f_(d)` are related by following formula

`(1)/(f_(d)) =(1)/(v)- (1)/(u)`.............(2)

put values of
`f_(d)` and u in equation (2)

`(1)/(-12) =(1)/(v)- (1)/(-2)`

`(1)/(-12)-(1)/(2) =(1)/(v)`

`(-2-12)/(2* 12) =(1)/(v)`

`v=(2* 12)/(-14)\\ v=-1.71 cm`

It means final image is located at 1.71 cm to the left of diverging lens. Final image will be inverted and virtual.