An object is placed 14.0 cm from a first converging lens of focal length 11.5 cm

Question

An object is placed 14.0 cm from a first converging lens of focal length 11.5 cm. A second converging lens with focal length 5.00 cm is placed 10.0 cm to the right of the first converging lens. (a) Find the position q1 of the image formed by the first converging lens. cm (b) How far from the second lens is the image of the first lens? cm beyond the second lens (c) What is the value of p2, the object position for the second lens? cm (d) Find the position q2 of the image formed by the second lens. cm (e) Calculate the magnification of the first lens. (f) Calculate the magnification of the second lens. (g) What is the total magnification for the system?

Explanation / Answer

(a)

lens formula:

1/v+1/u=1/f

1/v+1/14=1/11.5

1/v=5/322

v =64.4 cm

(b)

Distance = 64.4-10 =54.4 cm

(c)

p2 = - 54.4 cm

object position sign is negative as this image is behind the second lens.

(d)

1/v-1/54.4 = 1/5

1/v = 297/1360

v = 4.579 cm

(e)

M1 = -64.4/14 = -4.6

(f)

M2 =4.579/54.4 = 0.0842

(g)

M = M1*M2 = -0.38732

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