Question

A $4\,cm$ tall object is placed on the principal axis of a convex lens. The distance of the object from the optical centre of the lens is $12\,cm$ and its sharp image is formed at a distance of $24\,cm$ from it on a screen on the other side of the lens. If the object is now moved a little away from the lens, in which way (towards the lens or away from the lens) will he have to move the screen to get a sharp image of the object on it again? How will the magnification of the image be affected?

Answer 1

Hint: Here we have to first use the Lens formula to find the focal length.
Then we can find the magnification using the object distance and the image distance. Using the focal length and object distance we can find what happens to the image when the object is moved away from the lens.

Complete step by step answer:
Applying the lens formula we get:
Let us consider $v$ as the distance of the image from the screen on the other side of the lens, $u$ as the distance of the object from the optical centre of the lens, $f$ as the focal length.
$
 \dfrac{1}
{v} - \dfrac{1}
{u} = \dfrac{1}
{f} \\
 \Rightarrow \dfrac{1}
{f} = \dfrac{1}
{{24}} - \dfrac{1}
{{ - 12}} \\
 \Rightarrow \dfrac{1}
{f} = \dfrac{1}
{8} \\
 \Rightarrow f = 8 \\
$
When $u = - 12$ , the object is located between $f$ and $2f$ and the image is created beyond $2f$ on the other side. This is an enlarged inverted image. When the object is shifted away from the lens, the image turns toward the lens. Then the screen should be shifted towards the lens.
The magnification is given by:
$
 m = \dfrac{v}
{u} \\
 = \dfrac{{24}}
{{ - 12}} \\
= - 2 \\
$
Since, magnification is indirectly proportional to the object distance, the magnification decreases.

Additional information:
- Although virtual images are often created by diverging lenses, converging lenses are able to generate both actual and virtual images. When the target is positioned less than one focal length from the converging lens, a virtual image is created.
- Any incident ray passing parallel to a converging lens’s main axis will refract through the lens and pass on the opposite side of the lens through the focal point.
- Any incident ray passing on the way to the lens through the focal point will refract through the lens and travel parallel to the main axis.
- In effect, an incident ray that travels through the lens centre would continue in the same direction it had when it reached the lens.

Note:
Here the object distance is taken as negative since the image will be formed on the other side of the lens. We have to carefully observe what is provided in the question. At the middle, a convex lens is thicker than at sides. A convex lens is a lens that converges. As parallel light rays travel through a convex lens, at one point called the principal focus, the refracted rays converge. The focal length is considered the interval between the principal focus and the centre of the lens