Question

A $16cm$ long pencil is placed as shown in figure. The central point C is $45cm$ away from a $20cm$ focal length lens and $15cm$ above the optic axis. Find the length of the image.

Answer 1

Hint : Here we are asked to find the length of the image of the pencil $A'B'$ . For this, we will first find the coordinates of all the three given points for its image. We will use the lens formula for finding these coordinates.

Complete Step By Step Answer:
As we are given the position of the midpoint of the pencil, we will first find the coordinates for the image of this midpoint $C'$ .
For point C, the object distance $u = - 45cm$ .
We know that the lens formula is: $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ , where, $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the lens.
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{{20}} + \dfrac{1}{{ - 45}} = \dfrac{{9 - 4}}{{180}} = \dfrac{5}{{180}} = \dfrac{1}{{36}} \\\ \Rightarrow v = 36cm \\\$
We know that height of the image is given by: $h' = \dfrac{v}{u} \times h$ , where, $h'$ is the height of the image, $v$ is the image distance, $u$ is the object distance and $h$ is the height of the object.
We are given $h = 15cm$ for C.
$\Rightarrow h' = \dfrac{{36}}{{ - 45}} \times 15 = - 12cm$
Therefore, the coordinates of $C'$ are $\left( {36, - 12} \right)$ .
Now, we will do a similar process for point B.
For B, $u = - \left( {45 + 8\cos 45} \right) = - 50.66cm$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{{20}} + \dfrac{1}{{ - 50.66}} = \dfrac{1}{{33}} \\\ \Rightarrow v = 33cm \\\$
The height of the point $B'$ will be
$h' = \dfrac{{33}}{{ - 50.66}} \times \left( {15 - 4\sqrt 2 } \right) = - 6.1cm$
Therefore, the coordinates of $B'$ are $\left( {33, - 6.1} \right)$ .
For point A, $u = - \left( {45 - 8\cos 45} \right) = - 39.34cm$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{{20}} + \dfrac{1}{{ - 39.34}} = \dfrac{1}{{40.7}} \\\ \Rightarrow v = 40.7cm \\\$
The height of the point $A'$ will be
$h' = \dfrac{{33}}{{ - 39.34}} \times \left( {15 + 4\sqrt 2 } \right) = - 21.4cm$
Therefore, the coordinates of $A'$ are $\left( {40.7, - 21.4} \right)$ .
Thus, the length of image of the pencil $A'B' = \sqrt {{{\left( {40.7 - 33} \right)}^2} + {{\left( { - 21.4 - \left( { - 6.1} \right)} \right)}^2}} = \sqrt {{{7.7}^2} + {{3.1}^2}} = 17.1cm$ .

Note :
In this question, we have used the concept of finding the distance between two points for determining the length of the image. Let us consider that we are given two points $\left( {a,b} \right)$ and $\left( {c,d} \right)$ . Then, the distance between these two points is given by $\sqrt {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}}$ .