The Problems:
1) The objective lens of an astronomical telescope has a focal length of 6 ft. The eyepiece has a focal length of 2 inches.
Find the angular magnification that the telescope will produce when used for distant objects.
(2) A telephoto lens consists of a converging lens of focal length 6 cm placed 4 cm in front of a diverging lens of focal length (-2.5 cm).
a) Do a graphical construction of the system showing where the image would be.
b) Compare the size of the image formed by this combination with the size of the image that would be formed by the positive lens alone.
Solutions:
(1) The magnifying power is defined by:
m = F/f(e)
where F is the focal length of the objective (the main or front lens) and f(e) is the focal length of the eyepiece.
We have F = 6 ft. = 6 x 12 in. = 72 in.
f(e) = 2 in.
Then: m = F/f(e) = 72 in./ 2 in. = 36x
(2) The graphical construction (diagram) for the problem is shown below:
The image AB that would have been formed by the converging (positive) lens alone is:
(6 cm - 4 cm) = 2 cm beyond the f = (-2.5 cm) lens and is taken as the virtual object for that lens. Then: s1 = -2 cm, and:
1/s1' = 1/f - 1/s1 = 1/ (-2.5) - 1/(-2) = -1/2.5 + 1/2 = 1/10
Then: s1' = 10 cm
Thus, the final image A'B' is real and 10 cm beyond the diverging lens - as the graphical construction shows.
The linear magnification: M1 = (-s1'/ s1) = (10 cm/ 2 cm) = 5
and since, h'/h = 5, then h' = 5h so the image formed by the combination is 5x larger than that formed by the (+) lens alone.
No comments:
Post a Comment