Answer:
image is located at 22.5cm to the left of diverging lens.
Explanation:
In this question we have given,
object distance from converging lens,u=-16cm
focal length of converging lens,[tex]f_{c}=12cm[/tex]
focal length of diverging lens,[tex]f_{d}=-10cm[/tex]
distance between converging and diverging lens=20cm
we have to find location of final image,v=?
Here we will first find the location of image formed by converging lens
we know that u, v and [tex]f_{c}[/tex] are related by following formula
[tex]\frac{1}{f_{c}} =\frac{1}{v}- \frac{1}{u}[/tex].............(1)
put values of [tex]f_{c}[/tex] and u in equation (1)
we got,
[tex]\frac{1}{12} =\frac{1}{v}- \frac{1}{-16}[/tex]
[tex]\frac{1}{12}-\frac{1}{16} =\frac{1}{v}[/tex]
[tex]\frac{16-12}{16\times 12} =\frac{1}{v}[/tex]
or
[tex]v=\frac{16\times 12}{16-12}\\ v=48cm[/tex]
The image is located 48 cm to the right of the converging lens. This image is real and inverted.
This image will object for the diverging lens which is now located (48-20=18) cm at the right hand side of diverging lens
it means,
u=18cm
we know that u, v and [tex]f_{d}[/tex] are related by following formula
[tex]\frac{1}{f_{d}} =\frac{1}{v}- \frac{1}{u}[/tex].............(2)
put values of [tex]f_{d}[/tex] and u in equation (2)
[tex]\frac{1}{-10} =\frac{1}{v}- \frac{1}{18}[/tex]
[tex]\frac{1}{-10}+\frac{1}{18} =\frac{1}{v}[/tex]
[tex]\frac{-18+10}{10\times 18} =\frac{1}{v}[/tex]
or
[tex]v=\frac{10\times 18}{10-18}\\ v=-22.5cm[/tex]
It means image is located at 22.5cm to the left of diverging lens.
