Question

A equiconvex lens of focal length $20\text{cm}$ is cut into two equal parts to obtain two plano- convex lenses. The two parts are then put in contact as shown in figure. What is the focal length of combination?

• A. $\text{Z}ero$
• B. $5\text{cm}$
• C. $10\text{cm}$
• D. $20\text{cm}$

Answer 1

The correct option is D $20\text{cm}$

For an equiconvex lens,

$f=20\text{cm}$

For the plano convex lens,

$\because f_1=f_2=2f$

The combination can be seen as combination of two plano convex lens and one equiconcave lens.



Refractive index of equiconcave lens is 1, since it is filled with air.
Therefore focal length will be infinite, hence, $\frac{1}{\text{focal length}}$ will be zero.

$\therefore \frac{1}{f_{net}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{1}{2f}+0+\frac{1}{2f}=\frac{1}{f}$

$\Rightarrow f_{net}=20\text{cm}$

Note: focal length of lens will not change even if the orientation of the lens is changed as long the radii of curvatures and RI are same.