An object is placed 16.4 cm from a first converging lens of focal length 10.8 cm
ID: 1519755 • Letter: A
Question
An object is placed 16.4 cm from a first converging lens of focal length 10.8 cm. A second converging lens with focal length36 cm is placed 10 cm to the right of the first converging lens. (Take the direction to the right to be positive.)
(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
(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.
Your response differs from the correct answer by more than 100%.
(e) Calculate the magnification of the first lens.
M1 =
(f) Calculate the magnification of the second lens.
M2 =
Your response differs from the correct answer by more than 100%.
(g) What is the total magnification for the system?
Mtotal =
(h) Is the final image real or virtual? Is it upright or inverted? virtualuprightrealinvertedno image
Explanation / Answer
Given
Focal length of first converging lens f 1 = 10.8 cm
Focal length of second converging lens f 2 = 36 cm
Object distance of first lens p1 = 16.4 cm
Distance between two lenses d = 10 cm
a)
Image formed by first converging lens is
1 / p1 + 1 / q 1 = 1/ f 1
1 / q 1 = 1 / f 1 - 1 / p 1
= 1 / 10.8 cm - 1 / 16.4 cm
= 0.02377
q 1 = 31.61 cm
_____________________________________________
b)
Disatnce between image of first lens and second lens is
p2 = q 1 - d
= 31.61 cm - 10 cm
=21.61 cm
_______________________________________________
c)
object position for the second lens
p2 = 21.61 cm
_______________________________________________
d)
Image formed by the second lens.
1 / p2 + 1 / q 2 = 1 / f 2
1 / q 2 = 1 / f 2 - 1 / p2
= 1 / 36 cm + 1 / 21.61 cm
q 2 = 7.4 cm
_______________________________________________
e) Magnification of the first lens
m 1 = - q 1 / p 1
= - ( 31.61 cm ) / ( 16.4 cm )
= - 1.92
_______________________________________________
f) Magnification of second lens is
m 2 = - q 2 / p 2
= - 7.4 / 21.61
= - 0.342
_______________________________________________
g)
Total magnification for the system is
M = m 1 m2
= ( - 1.92 ) ( -0.342 )
= 0.657
Final image is virtual image .
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