Question

What is seen on the screen when the flame is very close (at about $5cm$) to the lens?

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:
To begin with, we understand the lens formula for a convex lens which is given by the equation,
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.
Now, as we move the lens towards the object to focus on the screen, it is also found out that the image gets magnified as the $u$ distance reduces and the intensity of the light rays formed are lesser. Now, in this case, it is given that the source is moved very closer to the lens, i.e. about $5cm$ to the lens.
In this scenario,
\[\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}\]
$f$ is known to be 10 cm for the given lens, therefore
\[\dfrac{1}{v} = \dfrac{1}{{10}} - \dfrac{1}{5}\]
Taking LCM of 10 and 5
\[\dfrac{1}{v} = \dfrac{{1 - 2}}{{10}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{10}}\]
\[ \Rightarrow v = - 10cm\]
So, in this case, we find the value of the v to be negative which implies that the image is formed on the same side as the source. In a convex lens, this condition isn’t possible and hence it is proven that the image will be virtual and won’t appear on screen.
Thus, the image won’t be formed on screen when it is moved lesser than the focus point of the lens.

Note:
Virtual Images are possible, when the light rays passing through the focus and the lens meet at a given point within the aperture and generally lie on the same side of the object, which makes it difficult to catch it on a screen. These types of images are formed 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.