Question

The figure shows an equiconvex lens of focal length f. If the lens is cut along PQ, the focal length of each half will be :

• A. $\frac{f}{2}$
• B. $f$
• C. $2f$
• D. $4f$

Answer 1

The correct option is B $f$

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

$f=\frac{R}{2(\mu -1)}$

along PQ radius of will be infinte

$\frac{1}{f_r}=(\mu -1)\left(\frac{1}{R}\right)$

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

$f_r=f$