Question

A student focuses the image of a candle flame, placed at about $2m$ from a convex lens of focal length $10cm$, on a 2 screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.
In which direction does he move the lens to focus the flame on the screen?
What happens to the size of the image of the flame formed on the screen?
What difference is seen in the intensity (brightness) of the image of the flame on the screen?
What is seen on the screen when the flame is very close (at about $5cm$) to the lens?

Answer 1

For solving this question, we need to know characteristics of the images formed by the convex lens when an object is placed at various distances. Here, we will first use the lens formula to know the direction in which the lens should be moved to focus the flame on the screen. Then, by using the magnification formula for a convex lens, we will determine the size of the image and its intensity. Then again by using the lens formula, we will find the answer to the last question.

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:
Here, we are given that candle is placed at about $2m$ from a convex lens of focal length $10cm$,
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\\
But, $u$is negative
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
In the question, it is given that the student moves gradually the flame towards the lens and each time focuses its image on the screen. Therefore, $u$ decreases and $\dfrac{1}{u}$ increases. As per the formula, if $\dfrac{1}{u}$increases, $\dfrac{1}{v}$ will decrease. And hence, $v$ increases. Thus, our first answer is: The lens is moved away from the screen and towards the object to focus the flame on the screen.

Now, we know that magnification of a convex lens is given by $m = \dfrac{v}{u}$.We have seen that the object distance is decreased and image distance is increased. Thus, magnification will be increased.Hence, our second answer is: The size of the image of the flame formed on the screen is increased.

The image is enlarged therefore the light rays spread out.Thus, our third answer is: The intensity (brightness) of the image of the flame on the screen is increased.
For the final answer, we will again apply the lens formula.
\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\\ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} \\\
We have $f = 10cm$ and $u = - 5cm$
\dfrac{1}{v} = \dfrac{1}{{10}} + \dfrac{1}{{ - 5}} = - \dfrac{1}{{10}} \\\ \Rightarrow v = - 10cm \\\
Thus the image is virtual.
Also $m = \dfrac{v}{u} = \dfrac{{ - 10}}{{ - 5}} = 2$
Thus, the image is erect.Hence our fourth answer is: A virtual and erect image is seen on the screen when the flame is very close (at about $5cm$) to the lens.

Note: In this type of question we need to use two formulas. First, when the distance and nature of the image is to be determined, the lens formula is used. Second, when the size of the image is to be determined, the magnification formula is used.