Object O 1 is 16.5 cm to the left of a converging lens with a 11.2-cm focal leng
ID: 1316295 • Letter: O
Question
Object O1 is 16.5 cm to the left of a converging lens with a 11.2-cm focal length. A second lens is positioned 10.0 cm to the right of the first lens and is observed to form a virtual image at the position of the original object O1.
(a) What is the focal length of the second lens?
cm
(b) What is the overall magnification of this system?
(c) What is the nature (i.e., real or virtual, upright or inverted) of the final image?
real and invertedvirtual and inverted real and uprightvirtual and upright
Explanation / Answer
s1 = 16.5
s1' = ?
1/s1 + 1/s1' = 1/f1
(1/16.5)+(1/s1') = (1/11.2)
s1' = 35 cm
M1 = -s1'/s1 = -2.12
for lens 2
s2 = -(35-10) = -25 cm
s2' = -(16.5+10) = -26.5 cm
1/s2 + 1/s2'= 1/f2
-(1/25) - (1/26.5) = 1/f2
f2 = 12.86 cm
M2 = -s2'/s2 = -26.5/25 = -1.06
b)
M = M1 * M2 = +2.2472
c)
virtual and upright
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