An object is placed 17.2 cm from a first converging lens of focal length 10.8 cm

Question

An object is placed 17.2 cm from a first converging lens of focal length 10.8 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)?

realvirtual    

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

uprightinverted   

Explanation / Answer

(a) first image's position:
1/f1 = 1/q1 + 1/p1
1/q1 = 1/10.8 - 1/17.2 = 40/1161
q1 = 1161/40 cm = 29.02 cm (real image)

(b) 29.02 cm + 10 cm = 39.02 cm

(c) p2 = 39.02 cm

(d) second image's position:
1/f2 = 1/q2 + 1/p2
1/q2 = 1/5 - 1/39.02 = 1701/9755
q2 = 9755/1701 cm = 5.7 cm (virtual image)

(e) M1 = -q1/p1 = 29.02/17.2 = 1.68

(f) M2 = -q2/p2 = 5.7/39.02= 0.14

(g) M = M1 * M2 = 25/4

(h) real and inverted image

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