An object is placed 10 m before a convex lens with focal length 5 m . Another co

Question

An object is placed 10 m before a convex lens with focal length 5 m . Another concave lens is placed 17.5 m behind the first lens with a focal length ?8.1 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer.

Part 1

At what distance is the first image from the first lens? Answer in units of m.

Part 2

What is the magnification of the first image?

Part 3

At what distance is the second image from the second lens? Answer in units of m.

Part 4

What is the magnification of the final image, when compared to the initial object?

Explanation / Answer

Part1: For convex lens:
u = 10 m
f = 5.0 m
v =?
1/v = 1/f - 1/u = 1/5.0 - 1/10
=> v = 10.0 m
Part 2: m = v/u = 10.0/10 = 1.0
Part 3: For concave lens:
u = [17.5 - 10.0] = 7.5 m
f = - 8.1 m
v =?
1/v = 1/f - 1/u = - 1/8.1 -1/ 7.5   
=> v = - 3.89 m (virtual image)
But since the object for concave lens is not real, the image will be real.

Part 4: m = 3.89/7.5 = 0.52
Combined magnification is 1.0*0.52 = 0.52
Final image is nearly 0.52 times the original size of the object.
====================================
The final image is at a distance of 17.5+3.89 = 21.39 m from the first lens.

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