Question

An object $5cm$ in length is held $25cm$ from a converging lens of focal length $10cm$ .Draw the ray diagram and find the position, size, nature of the image formed.

Answer 1

Hint As the height, distance of the object and the focal length of the object is given, use the distance formula and find the distance of the image and then calculate the magnification of the lens and calculate the height of the image and discuss the nature of the image.

Complete Step by step solution
Here the height of the object is $5cm$ and the distance of the object from the focal point is $25cm$, the focal length of the lens is $10cm$ .
We know the image distance formula, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ [where $v$is the position of the image, $u$is the distance of the image from focal length, $f$ is the focal length. ]
Now, $u = - 25$ and $f = 10$
Hence, $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$
Now we will substitute the given values as,
$ \Rightarrow \dfrac{1}{v} - \dfrac{1}{{ - 25}} = \dfrac{1}{{10}}$
After simplification we will get,
$ \Rightarrow \dfrac{1}{v} = \dfrac{3}{{50}}$
$ \Rightarrow v = 16.66$
Hence the distance of the image is $16.66cm$.
Now we know the magnification is the ratio of distance of image and distance of object.
Hence, $m = \dfrac{v}{u} = \dfrac{{16.66}}{{ - 25}} = - 0.66$
Similarly, we know the magnification is the ratio between the height of the image and height of the object.
$m = \dfrac{{{h_1}}}{{{h_2}}} \Rightarrow - 0.66 = \dfrac{{{h_1}}}{5} \Rightarrow {h_1} = - 3.3$
$\therefore {h_1} = - 3.3cm$
Hence the height of the object is $3.3cm$ and the negative sign shows that the image is inverted.
Hence, we can say the image is formed at $16.66cm$ on the opposite sides of the lens.
Now the nature of the image is real and inverted.

Now we will draw the ray diagram of the image:

Note
As we know that two rays are used to draw the ray diagram while following the laws of reflection, it should be kept in mind that the process of constructing a ray diagram is the same for all the positions of objects while the result of the ray diagram is different.