Question

The equiconvex lens has focal length 'f'. If it is cut perpendicular to the principal axis passing through optical centre, then focal length of each half is.

• A. $f_{/2}$
• B. f
• C. $3f_{/2}$
• D. $2f$

Answer 1

Answer: (D)

Len's Maker's formula:-

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

For equi-convex lens:-

$\frac{1}{f}=(\mu -1)(\frac{1}{R}-\frac{1}{-R})$

$⟹\frac{1}{f}=(\mu -1)\frac{2}{R}$

For, the plano-convex lens formed:-

$\frac{1}{f^{{}^{\prime }}}=(\mu -1)(\frac{1}{R_1}-\frac{1}{\infty })$

$\frac{1}{f^{{}^{\prime }}}=(\mu -1)\frac{1}{R}$

$⟹f^{{}^{\prime }}=2f$

Answer-(D)