An object is placed 10 cm to the left of a diverging lens (lens 1) of focal leng

Question

An object is placed 10 cm to the left of a diverging lens (lens 1) of focal length -10 cm. The lens is located 20 cm from a converging lens of focal length 50cm. Finally, a converging mirror with a radius of curvature of 100 cm is placed to the right of the last lens (see diagram).

a) what is the location of the final image after light rays have passed through each lens twice?

b) is this a virtual or real image?

c) what is the magnification of the final object?

s An object is placed 10 cm to the left of a diverging lens (ens 1) of focal length -10 cm. The is located 20 cm from a converging lens of length 50 cm. a) Finally a converging mirror with a radius of of 100cm is to the right of the last lens. What is the location of the final image after light rays have passed through each lens twice? b) Is this a virtual or real image? c) What is the magnification of the final image? d) Is the image upright or inverted? Lens 1 Lens 2 f-50cm 10 cm 20 cm

Explanation / Answer

for diverging lens

object distance p1 = 10 cm

focal length f1 = -10 cm

image distance q1 = ?

1/p1 + 1/q1 = 1/f1

1/10 + 1/q1 = -1/10

q1 = -5 cm

magnification m1 = -q1/p1 = 5/10 = 0.5

for converging lens

object distance p2 = 20 + 5 = 25 cm

focal length f2 = 50 cm

image distance q2 = ?

1/p2 + 1/q2 = 1/f2

1/25 + 1/q2 = 1/50

image distance q2 = -50 cm

magnification m2 = -q2/p2 = 50/25 = 2

for mirror

object distance p3 = 50 + 50 = 100 cm

focal length f = R/2 = 100/2 = 50 cm

image distance q3 = ?

1/p3 + 1/q3 = 1/f3

1/100 + 1/q3 = 1/50

q3 = 100 cm

magnification m3 = -q3/p3 = -100/100 = -1

image is 100 cm from the mirror

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(b)

real

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(c)

total magnification M = m1*m2*m3 = 0.5*2*-1 = -1

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(d)

inverted

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