An object is placed 10 m before a convex lens with focal length 5.4 m . Another

Question

An object is placed 10 m before a convex lens with focal length 5.4 m . Another concave lens is placed 8.61 m behind the first lens with a focal length ?2.2 m (see the figure below). Note: Make a ray diagram sketch in order to check your numerical answer.

1.) At what distance is the first image from the first lens? (Answer in meters)

2.) What is the magnification of the first image?

3.) At what distance is the second image from the second lens?

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

Explanation / Answer

Part 1)

Apply 1/f = 1/p + 1/q

1/5.4 = 1/10 + 1/q

q = 11.7 m behind the first lens

Part 2)

M = -q/p

M = -11.7/10 = -1.17 (Negative means its inverted)

Part 3)

Since the concave lens is 8.61 m behind the the first lens and the original image is 11.7 cm behind, that is 3.13 m behind the second lens and thus a virtual object, so...

1/-2.2 = 1/-3.13 + 1/q

q = -7.41

That is 7.41 m in front of the second lens

Part 4)

M for that part is -(-7.41)/-3.13

M = -2.37

Total M = -2.37 times -1.17

M = 2.77

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