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Question
An object of height 4 cm is kept at a distance of 30 cm from a concave lens. Use lens formula to determine the image distance, nature and size of the image formed if focal length of the lens is 15 cm.
Solution
We have
Height of object, h1= 4 cm
Focal length of lens, f = -15 cm
Object distance, u = -30 cm
Using lens formula, we have
`1/v-1/u=1/f`
`=>1/v-1/(-30)=1/(-15)`
`=>1/v+1/30=-1/15`
`=>1/v=-1/30-1/15`
`=>1/v=(-1-2)/30`
`=>1/v=-3/30`
`=>v=-30/3=-10` cm
Magnification, M`=v/u=-10xx(-1/30)`
`=1/3=0,.33`
Again, Magnification, M`=v/u=h_2/h_1`
`=>h_2/4=1/3`
`=>h_2=4/3=1.33` cm
Thus the image will be formed in front of the lens at a distance of 10 cm from the lens, virtual and erect of size 1.33 cm.
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