Question

i) Find the nature, position and magnification of the image formed by a convex lens of focal length $10cm$, if the object is placed at a distance of a) $15cm$ and b) $8cm$.
ii) Which of the above represents the use of a convex lens in a) film projector b) the magnifying glass used by a palm reader?

Answer 1

Hint:To solve this problem, first we need to use the lens formula for convex lens which gives the relation between its focal length, object distance and image distance. By determining the image distance from this formula, we can get the position of the image formed. After that we also need to use the formula for magnification to determine the nature of the formed image. Depending on all these details, we will be able to tell the suitable case applicable for given applications.

Formulas used:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$,
Where, $f$ is the focal length, $u$ is the object distance and $v$ is the image distance
$m = \dfrac{v}{u}$,
Where, $m$ is magnification, $u$ is the object distance and $v$ is the image distance

Complete step by step answer:
In the first question, we are given that focal length of convex length $f = 10cm$.
i) a) object distance $u = - 15cm$ which is between F and 2F
Now, applying the lens formula
$
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{1}{v} - \dfrac{1}{{\left( { - 15} \right)}} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{1}{v} + \dfrac{1}{{15}} \\
\Rightarrow \dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{{15}} \\
\Rightarrow \dfrac{1}{v} = \dfrac{{3 - 2}}{{30}} \\
\Rightarrow \dfrac{1}{v} = \dfrac{1}{{30}} \\
\Rightarrow v = 30cm \\
 $
From this value, we can say that the image is formed on the other side of lens at a distance of $30cm$
Now, we now that magnification is given by
\[m = \dfrac{v}{u} = \dfrac{{30}}{{ - 15}} = - 2\]

i) b) the object distance is given by $u = - 8cm$ which is between centre of lens and F.
Applying the lens formula,
$
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{1}{v} - \dfrac{1}{{\left( { - 8} \right)}} \\
\Rightarrow \dfrac{1}{{10}} = \dfrac{1}{v} + \dfrac{1}{8} \\
\Rightarrow \dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{8} \\
\Rightarrow \dfrac{1}{v} = \dfrac{{4 - 5}}{{40}} \\
\Rightarrow \dfrac{1}{v} = - \dfrac{1}{{40}} \\
\Rightarrow v = - 40cm \\
 $
From this value, we can say that the image is formed on the side as the object at a distance of $40cm$.
Now magnification \[m = \dfrac{v}{u} = \dfrac{{ - 40}}{{ - 8}} = 5\]
This positive magnification indicates that the image formed is virtual, erect and of enlarged size.

Now, in the second question, two applications of convex lens are given:
ii) a) One is the film projector in which the convex lens having object distance $15cm$ is used because in the projector, the image should be produced on the other side of the lens.

ii) b) And the other is the magnifying glass used by palm readers in which the convex lens having object distance $8cm$ is used because in this, the image should be at the same side as that of the lens and it should be of large size.

Note:We know that convex lens is also called a converging lens or positive lens and is thicker in the middle. The light rays that pass through a convex lens converge or are brought closer together. Here we have seen two important applications of convex lenses which are in the film projector and in magnifying glasses. Moreover, the human eye, camera, microscope and telescope are also examples of convex lens.