Question

A concave lens has a focal length of $20\;cm$ . At what distance from a lens a $5\;cm\;$ tall object should be placed, so that it forms an image at $15\;cm\;$ from the lens? Also, calculate the size of the image formed.

Answer 1

The image distance can be estimated with the knowledge of object distance and focal length with the help of the lens formula. In optics, the relationship between the distances of an image $\left( v \right)$, the distance of an object $\left( u \right)$, and the focal length $\left( f \right)$of the lens are given by the formula called as Lens formula. The lens formula is used for convex as well as concave lenses. Hence we can solve this problem by Lens formula.

Complete answer:
As given in the question,
Image distance, $v = - 15\;cm\;$ (negative because of the convex lens)
Focal length, $f = - 20\;cm$
Let the distance of the object $u$.
By the lens formula we get $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
which gives us $\dfrac{1}{{ - 15}} - \dfrac{1}{u} = \dfrac{1}{{ - 20}}$
Therefore, we get $u = - 60cm$
Hence the object is located $60\;cm\;$ away from the lens and is on the same side as the image.
Now the height of the object, ${h_1} = 5cm$
Magnification, $m = \dfrac{{{h_1}}}{{{h_2}}} = \dfrac{v}{u}$
Putting the values for v and u we get,
Magnification, $m = \dfrac{{{h_2}}}{5} = \dfrac{1}{4}$
Therefore ${h_2} = \dfrac{5}{4} = 1.25cm$
Hence the height of the image is $1.25cm$ and the positive sign indicates that the image is virtual and erect

Note:
The distances are measured from the pole of the mirror. According to the conventional method, the negative sign specifies the distance measured in the direction opposite to the incident ray while the positive sign specifies the distance measured in the direction of the incident ray. The distance below the axis is negative while the distance above is positive.