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