Question

An equiconvex lens of glass of focal length 0.1 m is cut along a plane perpendicular to principal axis into two equal parts. The ratio of focal length of new lens formed and original lens is

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. $2:\frac{1}{2}$

Answer 1

Answer: (C)

2 : 1

For the biconvex lens of radii of curvatures R.

The focal length would be given by

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

The lens appears as given in the figure after cutting.

The new focal length is given by-

$\frac{1}{f_{final}}=(\mu -1)(\frac{1}{R}-\frac{1}{\infty })$

Hence, $\frac{f_{final}}{f_{initla}}=2$