A converging lens (f= 12.4 cm) is located 30.3 cm to the left of a diverging len

Question

A converging lens (f= 12.4 cm) is located 30.3 cm to the left of a diverging lens (f=-6.24 cm). A postage stamp is placed 38.7 cm to the left of the converging lens. (a) Locate the final image of the stamp relative to the diverging lens. (Include sign to indicate which side of the lens the image is on.) cm (b) Find the overall magnification (c) Is the final image real or virtual? real viriisal (d) With respect to the original object, is the final image upright or inverted? O upright inverted (e) With respect to the original object, is the final image larger or smaller? larger smaller

Explanation / Answer

for converging lens 1

object distance s1 = 38.7 cm

focal length f1 = 12.4 cm

image distacne s1' = ?

1/s1 + 1/s1' = 1/f1

1/38.7 + 1/s1' = 1/12.4

s1' = 18.2 cm

m1 = -s1'/s1 = -18.2/38.7 = -0.47

the image at s1' will be thw object for lens 2

for lens 2
_____________

object distance s2 = 30.3 - 18.2 = 12.1 cm

focal length f2 = -6.24 cm

image distance s2' = ?

1/s2 + 1/s2' = 1/f2

1/12.1 + 1/s2' = -1/6.24

s2' = -4.12 cm

m2 = -s2'/s2

m2 = 4.12/12.1 = 0.34

part(a)

final image distance s2' = - 4.12 cm   <<<-------ANSWER

==========================

part (b)

overall magnification M = m1*m2

M = -0.47*0.34 = -0.16

overall magnification M = 0.16 <<<-------ANSWER

==========================

part(c)

virtual   <<<-------ANSWER

=======================

part(d)

m is negative

image is inverted   <<<-------ANSWER

=====================

(e)

M < 1

image is SMALLER <<<<-------ANSWER

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