A diverging lens has a focal length of magnitude 20.2 cm (a) Locate the images f

Question

A diverging lens has a focal length of magnitude 20.2 cm (a) Locate the images for each of the following object distances. 40.4 cm distance locationn cm Select- 20.2 cm distance locationn cm Select- 10.1 cm distance locationn cm Select- (b) Is the image for the object at distance 40.4 real or virtual? real virtual Is the image for the object at distance 20.2 real or virtual? real virtual Is the image for the object at distance 10.1 real or virtual? real virtual (c) Is the image for the object at distance 40.4 upright or inverted? upright inverted

Explanation / Answer

(a) 1/f= 1/do + 1/di

do= 40.4 cm, f= -20.2 cm (diverging lens)

1/-20.2 = 1/40.4 +1/di

-1/20.2 - 1/40.4 = 1/di

di=-13.47 cm . the image lies between optical centre and focus.

when the object distance= 20.2 cm

1/f= 1/do + 1/di

do= 20.2 cm, f= -20.2 cm (diverging lens)

1/-20.2 = 1/20.2+1/di

-1/20.2 - 1/20.2 = 1/di

di=- 10.1 cm . the image lies between optical centre and focus.

when the object distance= 10.1 cm

1/f= 1/do + 1/di

do= 20.2 cm, f= -10.1 cm (diverging lens)

1/-20.2 = 1/10.1+1/di

-1/20.2 - 1/10.1 = 1/di

di=- 6.73 cm . the image lies between optical centre and focus.

(b)when the object distance= 40.4cm, image is virtual.

when the object distance= 20.2cm, image is virtual.

when the object distance= 10.1cm, image is virtual.

(c) m= - di/do

when the object distance= 40.4 cm, di= -13.47

m= - (-13.47/40.4)=0.33

when the object distance= 20.2 cm, di= -10.1 cm

m= - (-10.1/20.2)=0.5

when the object distance= 10.1 cm, di= -6.73cm

m= - (-6.73/10.1)=0.67

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