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Question
A lens of focal length 12 cm forms an erect image three times the size of the object. The distance between the object and image is ______.
Options
8 cm
16 cm
24 cm
36 cm
Solution
A lens of focal length 12 cm forms an erect image three times the size of the object. The distance between the object and image is 16 cm.
Explanation:
Given:
Magnification, M = 3
Focal length f = 12 cm
Image distance v = ?
Object distance u = ?
We know that:
`M = v/u`
Therefore
`3 = v/u`
3u = v
Putting these values in the lens formula, we get:
`1/v-1/u = 1/f`
⇒ `1/(3u)-1/u = 1/12`
⇒ `(1-3)/(3u) = 1/12`
⇒ `(-2)/(3u) = 1/12`
⇒`3u = −24`
⇒ `u = (-24)/3`
⇒ u = −8 cm
v = 3u = −8 × 3 = −24 cm
The distance between the object and image is
∣v − u∣ = ∣−24 − (−8)∣ = ∣−24 + 8∣ = ∣−16∣ = 16cm.
The distance between the object and image is 16 cm.
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Reason: Magnification = `"image distance"/ "object distance"` = `(-"v"/"u")`
