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Question
If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image?
Solution
Object distance (u) =-3
`1/f=1/v-1/u`
⇒`1/8=1/v-1/-3`
⇒`1/8=1/v+1/3`
⇒`1/8-1/3=1/v`
⇒`(3-8)/24=1/v`
⇒`-5/24=1/
⇒`v=-24/5`
⇒`v=-4.8`cm
The image will be at a distance of 4.8 cm in front of the lens.
Magnification `(m) v/u`
⇒ `m=(-4.8)/-3`
⇒m=1.6
Positive value of magnifiction shows that the image is virtual and erect.
`m=h_i/h_o`
⇒`1.6=h_i/3`
⇒`h_i=3xx1.6=4.8` cm
positive sign of the image shows that the image will be formed above the principal axis.
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