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Question
An object is placed 20 cm from (a) a converging lens, and (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.
Solution
) Focal length of the converging lens, i.e., convex lens f = + 15 cm
Object distance u = - 20 cm
Using the lens formula:
`1/f=1/v-1/u`
⇒`1/15=1/v-1/-20`
⇒`1/15=1/v+1/20`
⇒`1/v=1/15-1/20`
⇒`1/v=(4-3)/60=1/60`
∴ v = +60 cm.
Therefore, the image formed is real. It is at a distance of 60 cm from the lens and to its right.
`"Magnification" ="image distance"/ "object distance"`
∴ `m=v/u=60/-20=-3`
The magnification is greater than 1. Therefore, the image is magnified.
The negative sign shows that the image is inverted.
(b) Focal length of the diverging lens, i.e., concave lens f = - 15 cm
Object distance u = - 20 cm
Using the lens formula:
`1/f=1/v-1/u`
⇒`1/-15=1/v-1/-20`
⇒`1/v=1/-15-1/20`
⇒`1/v=(-4-3)/60=-7/60`
∴ v = - 8.57 cm.
Therefore, the image formed is virtual. It is at a distance of 8.57 cm from the lens and to its left.
`"Magnification"="image distance"/"object distance"`
∴ `m=v/u=(-8.57)/-20=+0.42`
The magnification is less than 1. Therefore, the image is diminished. The positive sign indicates that the image is erect.
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