A converging lens L_1 and a diverging lens L_2, with focal length f_1 = 40 cm an

Question

A converging lens L_1 and a diverging lens L_2, with focal length f_1 = 40 cm and f_2 (unknown) respectively, are separated by a distance D = 1 m as shown below. An object AB is placed a distance 80 cm in front of L_1 (to die left of L_1). 1. Find the location and the nature of the image A' B' through L_1. 2. If the final image created by this optical system forms 10 cm in front of L_2, what is the focal length L_2? What is the nature of the final image? 3. What is the total magnification of this optical system? Lens L_2 is now moved closer to L_1 so that they are separated by a distance d = 40 cm. 4. Using the lens equation show that the image A" B" and L_1 are superimposed (i.e., are at the same location). 5. What is the total magnification of this optical system?

Explanation / Answer

(1) For a converging lens, we have

1 / dAB + 1 / dA'B' = 1 / f1

1 / dA'B' = 1 / (40 cm) - 1 / (80 cm)

dA'B' = 80 cm

The nature of the image through L1 will be real.

(2) For a converging lens, we have

1 / dCD + 1 / dC'D' = 1 / f2

1 / f2 = 1 / (1 cm) - 1 / (10 cm)

f2 = 1.11 cm

The nature of the image through L2 will be real.

(3) The total magnification of this optical system which will be given as :

we know that, Mtotal = (dA'B' / dAB) (dC'D' / dCD)

Mtotal = [(80 cm) / (80 cm)] [(10 cm) / (1 cm)]

Mtotal = 10

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