Question

There is an equiconvex lens of focal length of $20cm$. If the lens is cut into tow equal parts perpendicular to the optical axis, the focal length of each part will be

• A. 20 cm
• B. 10 cm
• C. 40 cm
• D. 15 cm

Answer 1

The correct option is C 40 cm

The focal length of equiconvex lens is 20 cm

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})forequiconvexlens{R}_1=-R_2=Rthus,\frac{1}{20}=(\mu -1)(\frac{1}{R}+\frac{1}{R})\Rightarrow R=40(\mu -1)$

when lens is cut into two equal parts perpendicular to optical axis

then,

$R_1=\infty ,R_2=R$

thus

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

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

by putting value of R we get,

$\frac{1}{f^{\prime }}=(\mu -1)\left(\frac{1}{40(\mu -1)}\right)\Rightarrow f^{\prime }=40cm$

new focal length of each part is 40 cm , becomes double

Option C is correct.