An object of hight 4.0 cm is placed at a distance of 30 cm from the optical centre O' of a convex lens of focal length 20 cm.Draw a ray diagram to find the position and size of the image formed.Mark optical centre O' and principal focus F' on the diagram.Also find the approximate ratio of size of the image to the size of the object.

Dear Student ,
Image result for image formation when an object is placed in between the focus and radius of curvature of a convex lens

In the ray diagram object size=AB=4 cm,  the focal length=CF=20cm, object distance =CB=u=30 cm, so the image at distance v is formed at:            1f=1v-1uor           120=1v-1-30or              1v=120-130=130or                v=60cmWe know that   IO=vusoIO=6030=21
in the diagram above C is optical centre.
Regards

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V 60&ratio2:1
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V=60 ratio is 2:1
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1/v - 1/u = 1/f
1/v -(- 1/30) = 1/20
1/v + 1/30 = 1/20
1/v = 1/20 - 1/30
1/v = 3/60 - 2/60
1/v = 1/60
v = 60 cm
Now for finding image height
v/u = himage/hobject
60/-30 = himage/4
-2 = himage/4
himage = -8 cm
The object should be placed between 2F1 and F1

 
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Sorry i forgot to mention the ratio . The ratio is 2 : 1
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What are you looking for?