Question

A diverging lens has a focal length of -14.0 cm. Locate the images for each of the following object distances. For each case, state whether the image is real or virtual and upright or inverted, and find the magnification.

**(a) 28.0 cm**

______________ cm

real, erect

real, inverted

virtual, erect

virtual, inverted

**magnification** ________________ X

**(b) 14.0 cm**

______________ cm

virtual, erect

virtual, inverted

**magnification ___________________** ?

**(c) 7.0 cm**

_______________ cm

real, erect

real, inverted

virtual, erect

virtual, inverted

**magnification __________________** ?

Answer #1

(a)

object distane s = 28 cm

focal length f = -14 cm

image distance s' = ?

1/s + 1/s' = 1/f

1/28 + 1/s' = -1/14

s' = -9.33 cm

image distance 9.33 infront of the lens

virtual

erect

magnification m = -s'/s = 0.33

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

(b)

object distane s = 14 cm

focal length f = -14 cm

image distance s' = ?

1/s + 1/s' = 1/f

1/14 + 1/s' = -1/14

s' = -7 cm

image distance 7 cm infront of the lens

virtual

erect

magnification m = -s'/s = 0.5

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

(c)

object distane s = 7 cm

focal length f = -14 cm

image distance s' = ?

1/s + 1/s' = 1/f

1/7 + 1/s' = -1/14

s' = -4.67 cm

image distance 4.67 infront of the lens

virtual

erect

magnification m = -s'/s = 0.67

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