The equiconvex lens has focal length f. If is cut perpendicular to the principal axis passin through optical centre, then focal length of each half is (1) f/2 (2) f (3) 3f/2 (4) 2f

Answer: (4)

Len’s Maker’s formula is given by

\(\begin{array}{l}\frac{1}{f} = \left ( n-1 \right )\left ( \frac{1}{R_{1}} – \frac{1}{R_{2}}\right )\end{array} \)

where

f is the focal length (half the radius of curvature)

n is the refractive index of the material used

R1 is the radius of curvature of sphere 1

R2 is the radius of curvature of sphere 2

For equiconvex lens

\(\begin{array}{l}\frac{1}{f} = \left ( n-1 \right )\left ( \frac{1}{R} – \frac{1}{-R}\right )\end{array} \)

\(\begin{array}{l}\frac{1}{f} = \left ( n-1 \right )\frac{2}{R}\end{array} \)

For, the plano-convex lens formed

\(\begin{array}{l}\frac{1}{f^{‘}}= \left ( n-1 \right )\frac{1}{R}\end{array} \)

⇒ f’ = 2f

While an equiconvex lens is cut parallel to the principal axis, the focal length remains equal and when the lens is cut perpendicular to the principal axis, the focal length is twice the original.

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