Question

A convex lens produces real image m times the size of the object. What will be the distance of the object from the lens?
A. $\left( {m - 1} \right)f$
B. $\left( {\dfrac{{m + 1}}{m}} \right)f$
C. $\left( {\dfrac{{m - 1}}{m}} \right)f$
D. $\dfrac{{m + 1}}{f}$

Answer 1

Use the formula for magnification of the lens for real image to determine the image distance. Use lens formula for lens to determine the object distance with given information.

Formula used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Here, f is the focal length, v is the image distance and u is the object distance.

Complete step by step answer: We know that the magnification of lens is the ratio of image distance to object distance. The magnification formula for convex lens is,
$M = \dfrac{{ - v}}{u}$

Here, v is the image distance and u is the object distance.

We have given, the size of the image is m times the size of the object. Therefore, the magnification of the lens is,
$M = \dfrac{{mu}}{u}$

Comparing the above two equations, we get,
$v = - mu$
We know the lens formula relating focal length, object distance and image distance,
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Here, f is the focal length.

Substitute $- mu$ for v in the above equation.
$\dfrac{1}{f} = - \dfrac{1}{{mu}} - \dfrac{1}{u}$
$\Rightarrow - \dfrac{1}{f} = \dfrac{1}{u}\left( {\dfrac{1}{m} + 1} \right)$
$\Rightarrow - \dfrac{1}{f} = \dfrac{1}{u}\left( {\dfrac{{m + 1}}{m}} \right)$
$\Rightarrow u = - \left( {\dfrac{{m + 1}}{m}} \right)f$

For convex lenses, the object distance is negative. Therefore,
$u = - \left( { - \left( {\dfrac{{m + 1}}{m}} \right)f} \right)$
$\Rightarrow u = \left( {\dfrac{{m + 1}}{m}} \right)f$

So, the correct answer is option (B).

Note: Students should remember some of the properties of convex lens and concave lens. For convex lenses, object distance is negative and image distance is positive. For concave lens, object distance and image distance both are negative. Focal length of convex lens is positive whereas focal length of concave lens is negative. The magnification of the lens is negative for a real image and positive for a virtual image.