Question

The focal length of a symmetric biconvex lens is $f$. If we cut the lens into two equal pieces as shown below, the focal length of each piece will be ___.

• A. $\frac{f}{2}$
• B. $2f$
• C. $f$
• D. $0$

Answer 1

Answer: (A)

Correct option = (B) 2f
According to lens makers formula,
$\frac{1}{f}$ =$\text{(n-1)(}$$\frac{\text{1}}{R_1}$-$\frac{\text{1}}{R_2}$$\text{)}$
where, $f$ is the focal length of the lens in the medium where it is kept and $n$ is the refractive index of the lens with respect to the medium. $R_1$ and $R_2$ are the radii of curvature of the two refracting surfaces of the lens. For biconvex symmetric lens, $R_1$ = -$R_2$ = $R$.
So focal length will be,
$\frac{\text{1}}{f}$ =$\text{(n-1)(}$$\frac{\text{1}}{R}$+$\frac{\text{1}}{R}$$\text{)}$
$⟹$ $\frac{\text{1}}{f}$ =$\text{(n-1)}$$\frac{\text{2}}{R}$ -----(i)

After cutting the lens vertically, one of the surface becomes flat, which can be considered as the part of a very large sphere with radius of curvature infinity. So for each piece focal length will be,
$\frac{\text{1}}{f_{new}}$ =$\text{(n-1)}$$(\frac{\text{1}}{R}-0)$
$⟹$ $\frac{\text{1}}{f_{new}}$ = $\frac{\text{1}}{2f}$ (from equation (i))
$⟹$ $f_{new}$ = 2f

Hence, focal length of each piece is 2f.