Optics and Maths

Classical Mechanics Level 3

The image of a small electric bulb fixed on the wall of a room is to obtained on the opposite wall 3 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?


The answer is 0.75.

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Keshav Tiwari by Keshav Tiwari , 23, India

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1 solution

Deepanshu Gupta
Feb 21, 2015

Let distance lens from bulb as ' x ' .

1 3 − x + 1 x = 1 f f = ( 3 − x ) ( x ) 3 ≥ 1 3 ( ( 3 − x ) + ( x ) 2 ) 2 = 3 4 ( ∵ A M ≥ G M ) f m a x = 0.75 \displaystyle{\cfrac { 1 }{ 3-x } +\cfrac { 1 }{ x } =\cfrac { 1 }{ f } \\ f=\cfrac { (3-x)(x) }{ 3 } \quad \ge \quad \cfrac { 1 }{ 3 } { (\cfrac { (3-x)+(x) }{ 2 } ) }^{ 2 }=\cfrac { 3 }{ 4 } \quad (\because AM\ge GM)\\ \boxed { { f }_{ max }=0.75 } }

8 Helpful 0 Interesting 0 Brilliant 0 Confused

You just love AM - GM . Don't you ? Btw Nice Solution !

Keshav Tiwari - 6 years, 3 months ago

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Yes I Love It ! It is Beautiful theorem.

Deepanshu Gupta - 6 years, 3 months ago

Can also be done at the last step by using D (>=) 0. Please correct me if I'm wrong.

9 - 12f (>=) 0.....therefore max value of f is 0.75 cm.

Ayan Jain - 6 years, 3 months ago

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Yes , you can do it either way .

Keshav Tiwari - 6 years, 3 months ago

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