1.

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

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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