An object is placed 15.6 cm from a first converging lens of focal length 10.2 cm

Question

An object is placed 15.6 cm from a first converging lens of focal length 10.2 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?
=____________

(h) Is the final image real or virtual (compared to the original object for the lens system)?

=__________

Is it upright or inverted (compared to the original object for the lens system)?
=__________

Explanation / Answer

a)

1/q = 1/f - 1/p = 1/10.2 - 1/15.6 >>>> q1 = 29.467 cm

b)

29.467-10 = 19.467 cm

c)
p2 = 19.467 cm

d)

1/q = 1/f - 1/-p = 1/5 + 1/19.467 >>>> q2 = 3.978 cm

The final image lies 3.978 cm to the right of lens 2.

e)

m1 = -29.467/15.6 = -1.89

f)

m2 = -3.978/-19.467 = 0.204

g)

m = m1*m2 = -1.89 * 0.204 = -0.386

h)
real

i)

upward

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