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An equiconvex lens of focal lenght 15 cm is cut into two halves of thickness. Find the focal length of each half.

+7 votes
asked Sep 9, 2016 in Physics by Rahul Roy (7,895 points) 31 96 261

1 Answer

+7 votes
answered Sep 9, 2016 by Abhishek Kumar (14,583 points) 5 9 34
selected Sep 22, 2016 by sarthaks
Best answer

From the lensmaker's equation, we have a symmetrical lens with R1 = -R2, and because it is a convex lens, R1>0. Then: 

1/(15 cm)= (n-1)*(1/R1-1/R2) 

1/(15 cm)= (n-1)*(2/R1) 

1/(30cm) = (n-1)/R1 

The new lenses produced by cutting the original lens in half (and we assume the cut is a flat surface) will have focal lengths: 

1/f = (n-1)*(1/R1 + 1/infinity) 
1/f = (n-1)*(1/R1 + 0) 
1/f = (n-1)/R1 

But we already showed that (n-1)/R1 = 1/(30 cm), so the focal length of each half of the lens is 30 cm.

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