Question

In which direction does he move the lens to focus the flame on the screen?

Answer 1

Hint: Understand the lens formula and apply the equation for the given case study question. Find out how the initial conditions are given for a convex lens and hence derive at your conclusion.

Complete Step By Step Solution:
In order to understand this question, we first analyze the lens formula,
Lens formula is given as,
\[\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\]
Where, $v$ is the final image distance, $u$ is the distance of the source from the lens and f is the focal length of the given lens.
It is given that the lens is convex. So, the image formed will be real and inverted in all cases. Hence, the image source is kept on the negative plane and the image forms on the positive x-axis. This implies, $u$ is negative and $v$ is positive.
Therefore, the equation changes as,
\[\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}\]
This implies that, when the source is moved near the lens, i.e. when it is moved near the lens, $\dfrac{1}{u}$ decreases, hence $\dfrac{1}{v}$ increases and hence $v$ increases. In this case, the image formed will not be clear, but rather blurred.
So, when u is decreased, $\dfrac{1}{u}$ is increased, which in turn increases the $v$ distance of the screen. Thus $\dfrac{1}{v}$ distance is reduced. So, the clear image is obtained when the source rays strike the focus point and thus it is required to move the lens towards the source.

Hence, the lens should be moved towards the source to obtain a clear image.

Note:
In a double concave or concave mirror, the image formed will be virtual and inverted or upright depending upon the position where the object is placed.