A converging lens (f= +20.0 cm) is located 60.0 cm to the left of a diverging le

Question

A converging lens (f= +20.0 cm) is located 60.0 cm to the left of a diverging lens (f= -30.0 cm). A 3.00 cm object is placed 10.0 cm to the left of the converging lens. (a) What is the position of the intermediate image (i.e., the image formed by lens one alone) relative to the converging lens? (b) What is the position of the final image relative to the diverging lens? (c) Is the final image real or virtual and is it upright or inverted with respect to the orientation of the original object? (d) What is the overall magnification of the object? (e) Provide a ray diagram of the situations.

Please show all work so I can follow with ease.

Explanation / Answer

for the lens f1 on left

a) 1/p1 + 1/q1 = 1/f1

1/10 + 1/q1 = 1/20

q1 = (10*20)/(10-20) = -20 cm

M1= -q1/p1 = 20/10 = 2

b) for right diverging lens

p2 = 60+20 = 80 cm

1/p2 + 1/q2 = 1/f2

1/80 + 1/q2 = -1/30

q2 = -21.81 cm

m2 = -q2/p2 = 21.81/80 = 0.273

c) virtual , upright

d) M = M1*M2 = 2*0.273 = 0.546

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