Question

A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principle axis. If the power of the original lens is $4D$, the power of a cut lens will be

• A. $3D$
• B. $4D$
• C. $5D$
• D. $2D$

Answer 1

Answer: (D)

As the convex lens is symmetric on both sides, i.e $R_1=R_2=R$

When the lens in not cut:

Using lens maker formula:

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

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

Power of lens, $P=\frac{1}{f}$

$\Rightarrow$ $4D=\frac{2}{R}(\mu -1)$

$\Rightarrow \frac{\mu -1}{R}=2D$ .........(1)

Now the lens in cut perpendicular to the principal axis:

$R_1=R$ and $R_2=\infty$

Using lens maker formula:

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

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

Power of lens, $P^{\prime }=\frac{1}{f^{\prime }}=$ $=\frac{1}{R}(\mu -1)=2D$ .......... using (1)