An object is placed 17 cmfrom a first converging lens of focal length 20 cm. A s

Question

An object is placed 17 cmfrom a first converging lens of focal length 20 cm. A second converging lens with focallength 37 cm isplaced 13.6 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? upright
real
noimage
inverted
virtual

(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? upright
real
noimage
inverted
virtual

upright
real
noimage
inverted
virtual

upright
real
noimage
inverted
virtual

Explanation / Answer

From lens formula 1/f = 1/p1+1/q1 1/q1 =1/f - 1/p1 = 1/20cm -1/17cm the image distance of the first lens q1= -113.3cm magnification m1 = -q1/p1 = 113.3/17 = 6.66 the image is at distance from the saecond lens =q1+L = 113.3+13.6 = 126.9cm the object distance of the second lens is p2 =126.9cm From lens formula 1/f2 = 1/p2+1/q2 1/q2 =1/f2 - 1/p2 =1/37cm - 1/126.9cm the image distance of the second lens q2 =52.2 magnification m2 =-q2 /p2  = -52.2/126.9 =-0.41 the overall magnification M = m1*m2 =6.66 * -0.41 = -2.7 so the final image is real and inverted 1/q1 =1/f - 1/p1 = 1/20cm -1/17cm the image distance of the first lens q1= -113.3cm magnification m1 = -q1/p1 = 113.3/17 = 6.66 the image is at distance from the saecond lens =q1+L = 113.3+13.6 = 126.9cm the object distance of the second lens is p2 =126.9cm From lens formula 1/f2 = 1/p2+1/q2 1/q2 =1/f2 - 1/p2 =1/37cm - 1/126.9cm the image distance of the second lens q2 =52.2 magnification m2 =-q2 /p2  = -52.2/126.9 =-0.41 the overall magnification M = m1*m2 =6.66 * -0.41 = -2.7 so the final image is real and inverted 1/f2 = 1/p2+1/q2 1/q2 =1/f2 - 1/p2 =1/37cm - 1/126.9cm the image distance of the second lens q2 =52.2 magnification m2 =-q2 /p2  = -52.2/126.9 =-0.41 the overall magnification M = m1*m2 =6.66 * -0.41 = -2.7 so the final image is real and inverted 1/q2 =1/f2 - 1/p2 =1/37cm - 1/126.9cm the image distance of the second lens q2 =52.2 magnification m2 =-q2 /p2  = -52.2/126.9 =-0.41 the overall magnification M = m1*m2 =6.66 * -0.41 = -2.7 so the final image is real and inverted

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