An object is placed 17 cmfrom a first converging lens of focal length 46 cm. A s

Question

An object is placed 17 cmfrom a first converging lens of focal length 46 cm. A second converging lens with focallength 50 cm isplaced 10.3 cm to the rightof the first converging lens. (a) Find the position q1 of the imageformed by the first converging lens.
cm

(b) How far from the second lens is the image of the firstlens?
cm

(c) What is the value of p2, the objectposition for the second lens?
cm

(d) Find the position q2 of the imageformed by the second lens.


(e) Calculate the magnification of the first lens.
M1 =

(f) Calculate the magnification of the second lens.
M2 =

(g) What is the total magnification for the system?
Mtotal =

(h) Is the final image real or virtual? Is it upright or inverted? virtual
inverted
upright
noimage
real

(a) Find the position q1 of the imageformed by the first converging lens.
cm

(b) How far from the second lens is the image of the firstlens?
cm

(c) What is the value of p2, the objectposition for the second lens?
cm

(d) Find the position q2 of the imageformed by the second lens.


(e) Calculate the magnification of the first lens.
M1 =

(f) Calculate the magnification of the second lens.
M2 =

(g) What is the total magnification for the system?
Mtotal =

(h) Is the final image real or virtual? Is it upright or inverted? virtual
inverted
upright
noimage
real

virtual
inverted
upright
noimage
real

virtual
inverted
upright
noimage
real

Explanation / Answer

(a) first image's position: 1/f1 = 1/q1 + 1/p1 1/q1 = 1/46 - 1/17 = -29/782 q1 = -782/29 cm = -27 cm (virtual image) (b) 27 cm + 10.3 cm = 37.3 cm (c) p2 = 37.3 cm (d) second image's position: 1/f2 = 1/q2 + 1/p2 1/q2 = 1/50 - 1/37.3 = -127/18650 q2 = -18650/127 cm = -146.9 cm (virtual image) (e) M1 = -q1/p1 = 27/17 = 1.6 (f) M2 = -q2/p2 = 146.9/37.3 = 3.9 (g) M = M1 * M2 = 25/4 (h) virtual and upright image

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