Question

A student focused the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his observation as given below:

S.no.	Distance of flame from lens $\left( cm \right)$	Distance of screen from the lens $\left( cm \right)$
1.	30	10
2.	20	12
3.	15	15
4.	12	20

Find out the length of the given lens.

Answer 1

Hint: The length of a lens refers to its focal length. The focal length is the distance of the focus from the pole of the mirror. Using the given values of observation and the lens formula and applying proper sign conventions, the focal length of the lens can very easily be found out.
The sign convention states that distances measured from the pole of the lens, measured opposite of the direction of the incident ray are negative and along the direction of the incident ray are positive.

Formula used:
The lens formula is as follows:
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ --(lens formula)
Where, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance. Proper sign convention should be followed while substituting the values.

Complete Step-by-Step solution:
The length of a lens refers to its focal length which is the distance of the focus from the pole of the mirror.
The values given to us in the table can be plugged in the lens formula after applying proper sign conventions to get the focal length of the lens.
The sign convention is as follows.
It states that the distances measured from the pole of the lens, measured opposite of the direction of the incident ray are to be considered as negative while those along the direction of the incident ray are to be considered as positive.
The object distance is always conventionally taken to be negative. Hence, $u$ is negative.
Since, a convex lens is a converging lens, it forms a real image on the opposite side of the lens (the side opposite to the one from which the incident ray is coming).
Hence, the image distance is always along the direction of incident ray and is positive. Hence, $v$ is positive.
The lens formula is as follows:
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$ --(lens formula)--(1)
Where, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance. Proper sign convention should be followed while substituting the values.
Now, with this information, we can find out the focal length using any of the observation values. Let us take observation 1.
Here, $v=10cm$,$u=-30cm$
Putting these values in the lens formula (1), we get,
$\dfrac{1}{f}=\dfrac{1}{10}-\left( \dfrac{1}{-30} \right)$
$\therefore \dfrac{1}{f}=\dfrac{1}{10}+\dfrac{1}{30}=\dfrac{3+1}{30}=\dfrac{4}{30}=\dfrac{1}{7.5}$
$\therefore f=7.5cm$
Hence, the focal length of the lens is $7.5cm$.

Note: The focal length could also have been found out without using the lens formula by careful inspection of the observation values. In observation 3, the image and object distance both are equal at $15cm$. This is only possible, when the object is at the center of curvature of the lens. Thus, we will get the distance of the center of curvature $\left( C \right)$ from the lens. The focal length $\left( f \right)$ is known to be just half of this distance, that is $f=\dfrac{C}{2}$. This result has in fact been derived from the lens formula for equal magnitude values of $v$ and $u$.
Thus, from observation 3, the focal length would have been $\dfrac{15}{2}=7.5cm$, which is the value that we earlier arrived at.