Question

A thin plano-convex lens of focal length $f$ is split into two halves. One of the halves is shifted along the optical axis. The separation between the object and image planes is $1.8m$. The magnification of the image formed by one of the half lenses is $2$.
Find the focal length of the lens used.

• A. $0.4m$
• B. $0.8m$
• C. $0.6m$
• D. $2m$

Answer 1

Answer: (A)

$0.4m$

Let magnification caused by the first lens be 2 and distance $O{L}_1=x$.

Distance v of image from first lens $L_1$ is given by

$m=\frac{v}{u}=2\Rightarrow v=2u=2x$.

Clearly, $u+v=1.8m\Rightarrow x+2x=1.8m$

or $3x=1.8m\Rightarrow x=\frac{1.8}{3}=0.6m$

By sign convention,

$u=-x=-0.6m,v=2x=1.2m$

Lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ gives

$\frac{1}{f}=\frac{1}{1.2}+\frac{1}{0.6}=\frac{1+2}{1.2}$

$\therefore$ Focal length $f=\frac{1.2}{3}=0.4m$

For real image, lens formula takes the form

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Clearly, u and v are interchangeable. Therefore, for lens $L_2$

$u^{\prime }=v=1.2m$ and $v^{\prime }=0.6m$

$O{L}_1=L_2{I}_2=x$

If d is the separation between the lenses, then

$x+d+x=1.8m$

$\therefore d=1.8-2x=1.8-2\times 0.6=0.6m$

Method-2 Since the magnification for $L_1$ is 2

$\Rightarrow \frac{v}{u}=-2\Rightarrow \frac{\frac{D+d}{2}}{-\frac{D-d}{2}}=-2$

$\Rightarrow \frac{D+d}{D-d}=2\Rightarrow D=1.8m,d=0.6m$.

$f=\frac{D^2-d^2}{4D}=\frac{(1.8+0.6)(1.8-0.6)}{4\times 1.8}=0.4m$