Question

A convex lens of focal length $f$ produces a virtual image $n$ times the size of the object. Then the distance of the object from the lens is:

• A. $(n-1)f$
• B. $(n+1)f$
• C. $\left(\frac{n-1}{n}\right)f$
• D. $\left(\frac{n+1}{n}\right)f$

Answer 1

The correct option is C $\left(\frac{n-1}{n}\right)f$


Given:

focal length $=f$
magnification, $m=\frac{h_i}{h_0}=n$

Applying lens formula for given condition,

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

$\Rightarrow \frac{u}{v}-1=\frac{u}{f}$

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

$\Rightarrow \frac{v}{u}=\frac{f}{u+f}.........\left(1\right)$

Now applying formula of magnification

$m=\frac{v}{u}=n$

$\Rightarrow \frac{f}{f+u}=n\left[\text{from eq.}\right(1\left)\right]$

$\Rightarrow f=nf+nu$

$f(1-n)=nu\Rightarrow u=(\frac{1-n}{n})f$

So, object distance (considering sign convention) for the given lens will be

$-u=\left(\frac{n-1}{n}\right)f$

Hence, option $\left(c\right)$ is correct.