In the figure are shown two converging lenses with x 2 = 1 m. The lens on the le
ID: 2120446 • Letter: I
Question
In the figure are shown two converging lenses with x2 = 1 m. The lens on the left (Lens 1) has a focal length of 0.25 m while the one on the right (Lens 2) has a focal length of 0.2 m. An object 2.9 cm high is located x1 = 0.419 m to the left of Lens 1.
Consider the image of the object formed by Lens 1. Give its distance as measured from Lens 1 and the image height.
Distance = .626
Height = -4.33
Now consider the image formed by Lens 1 as the object for Lens 2. Locate the image formed by Lens 2 as measured from Lens 2 and give the height of the image.
Distance = ??
Height = ??
I need help figuring out these last two questions.
Explanation / Answer
We have for a converging lens 1/f = 1/s1+1/s Then s1 = sf/(s-f) [since 1/f- 1/s=1/s1or1/s1= s-f/sf Then s1=sf/s-f)] =(0.419m*0.25m)/(0.419m-0.25m) =0.6198224852m Therefore the image formed by the first lens is 0.6198224852m fromthe first lens.
Therefore it is 1m-0.6198224852m = 0.38017751479m from the secondlens.
Therefore for second lens s =0.38017751479 focal length f =0.2m
then image distance s1 = (0.2m*0.38017751479m)/(0.38017751479m-0.2m) = 0.42200331871 m 0.42200331871 height of the image formed by the first lens =(2.9m)(0.619m)/(0.419m) =4.284248210023866cm
Then the height of the image formed by the second lens
=(4.284cm)(0.422m)/(0.380177)
=4.755279777577286cm