Question

A 4cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24cm. The distance of the object from the lens is 16cm. Find the position, size and nature of the image formed, using lens formula.

Answer 1

Hint: We will first find out the value of the distance of the image from the lens using the lens maker formula and then find out its magnification in order to find out the nature, size and position of the image formed. Refer to the solution below.
Formula used: $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, $m = \dfrac{v}{u}$

Step-By-Step answer:

As we look at the question, the given values are-
Focal length of the lens- $f = + 24cm$
The given distance of the object from the lens denoted by the syllabus $u = - 16cm$
As we can see that the distance of the object from the lens is smaller than the focal length of the lens. This will clear the fact that the image formed is between the focus and the center of curvature. In this above position of the object, the rays coming from the object will not meet on the other side of the lens at any point.
Thus, from the lens equation $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, we can find out the distance of the image from the lens-
$
   \Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\
    \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{u + f}}{{uf}} \\
    \\
   \Rightarrow v = \dfrac{{uf}}{{u + f}} \\
$
Putting the given values by the question in the above calculated equation, we have-
$
   \Rightarrow v = \dfrac{{uf}}{{u + f}} \\
    \\
   \Rightarrow v = \dfrac{{ - 16 \times 24}}{{ - 16 + 24}} \\
    \\
   \Rightarrow v = - 48cm \\
$
Here, the negative sign confirms that the image will be formed on the same side of the lens as the object is placed.
Now, the magnification will be found out by using the formula $m = \dfrac{v}{u}$. Thus,
$
   \Rightarrow m = \dfrac{v}{u} \\
    \\
   \Rightarrow m = \dfrac{{ - 48}}{{ - 16}} \\
    \\
   \Rightarrow m = + 3 \\
$
Since the magnification found is positive and bigger than 1, the image will be virtual, erect and magnified.

Note: A magnifying type of a lens is a typical example of the usage of this sort of lens. Whenever an object is positioned ahead of it at a distance less than the focal length of the lens, a magnified and upright version of the target is produced on the same side of the object. Hypermetropia or long-sightedness is also fixed using the lens.