Respuesta :
Answer:
Final image is located at 1.71 cm to the left of diverging lens. Final image will be inverted and virtual.
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,[tex]f_{c}[/tex]=6cm
distance between converging and diverging lens=18cm
focal length of diverging(concave) lens,[tex]f_{d}[/tex]=-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 [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}{6} =\frac{1}{v}- \frac{1}{-8}[/tex]
[tex]\frac{1}{6}-\frac{1}{8} =\frac{1}{v}[/tex]
[tex]\frac{8-6}{6\times 8} =\frac{1}{v}[/tex]
[tex]v=\frac{6\times 8}{2}\\ v=24cm[/tex]
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 [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}{-12} =\frac{1}{v}- \frac{1}{-2}[/tex]
[tex]\frac{1}{-12}-\frac{1}{2} =\frac{1}{v}[/tex]
[tex]\frac{-2-12}{2\times 12} =\frac{1}{v}[/tex]
[tex]v=\frac{2\times 12}{-14}\\ v=-1.71 cm[/tex]
It means final image is located at 1.71 cm to the left of diverging lens. Final image will be inverted and virtual.
