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Question
Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?
Solution
Object distance=u=-8cm
Focal length f = 10 cm
Image distance v=?
`1/"v" - 1/"u" = 1/"f"`
`1/"v" - 1/-8 = 1/10`
`1/"v" = 1/10 - 1/8`
`1/"v" = (4 - 5)/40`
`1/"v" = 1/-40`
v = - 40 cm
As the object is placed between the focus and optical center of the lens, the image formed is virtual and erect.
Magnification (m) = `v/u = (-0)/-8`
m = +5.0
The +ve sign shows that the image formed is erect.
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