ch, 23 C Chegg study IGuided solution x C secure https://www.webassign.net 15490

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ch, 23 C Chegg study IGuided solution x C secure https://www.webassign.net 15490668 7, -14.32 points SerCP923 23PO33 A diverging lens has a focal length of magnitude 22.4 cm. (a Locate the images for each of the fallowing object dista distal location -Select- 22.4 cm location -Select distance location Solect- the image fo e object distance 44.8 real or virtual? ge for the object at dista 224 virtual? ge for the object at distance 2 real or virtual? ge for the object at dista 44.8 uprig inverted? uprig inverted ge for the object at distal 22.4 upright verted? upngl e image for the object at distance 11.2 upright or vented? uprig inverted (d) Find the magnification for the object at distance 44-8 cm. ind the magnification for the object at distance 22.4 ind the magnification for the object at dista My Notes Ask Your Teacher

Explanation / Answer

focal length, f=-22.4 cm

a)

let,

object distance, u=44.8cm

image distance is v

use,

1/u + 1/v = 1/f

1/44.8 + 1/v = 1/(-22.4)

===> v=-14.93 cm

magnification m=-v/u

m=-(-14.93)/44.8 = 0.33

image is virtual, upright

b)

object distance, u=22.4cm

image distance is v

use,

1/u + 1/v = 1/f

1/22.4 + 1/v = 1/(-22.4)

===> v=-11.2 cm

magnification m=-v/u

m=-(-11.2)/22.4 = 0.5

image is virtual, upright

c)

object distance, u=11.2cm

image distance is v

use,

1/u + 1/v = 1/f

1/11.2 + 1/v = 1/(-22.4)

===> v=-7.47 cm

magnification m=-v/u

m=-(-7.47)/11.2 = 0.67

image is virtual, upright

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