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 ______.

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KollegeApply · Accepted Answer

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 $\Rightarrow$ 1/(3u)-1/u = $\frac{1}{12}$ $\Rightarrow$ (1-3)/(3u) = $\frac{1}{12}$ $\Rightarrow$ (-2)/(3u) = $\frac{1}{12}$ $\Rightarrow$3u = −24 $\Rightarrow$ u = (-24)/3 $\Rightarrow$ u = −8 cm v = 3u = −8 $\times$ 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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