Let the focal of the equiconvex lens be f and that of the plano-convex lens be f'.Let the radius of curvature for the equiconvex lens be R.
Applying Lens maker's formula
`1/f = (mu - 1)(1/R + 1/R) = (2(mu - 1))/R`
For the plano-convex lens, the focal length can be calculated as:
`1/f' = (mu - 1)(1/R) = (mu - 1)/R`
Thus, f' = 2f
Thus, the focal length of the plano-convex lens is twice that of the equiconvex lens.
Hence, the correct answer is option 2f.
A Convex Lens Of Focal Length F Is Cut Into Two Identical Parts Each Forming A Plano Convex Lens.What Is The Focal Length Of Each Part?
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