Question

At what distance should an object be placed from a convex lens of focal length \[18\;cm\] to obtain an image at \[24\;cm\] if on the other side. What will be the magnification in this case?

Answer 1

Hint:The thin lens formula relating the focal length, image distance and object distance needs to be applied to first find the distance at which the image is formed. The magnification that is required to be found out is given by a formula in terms of the image and object distance which has to be applied.

Formula used:
The thin lens formula for spherical lenses is given as:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
Where, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance.

Complete step by step answer:
The above problem revolves around the concept of refraction through lenses and the formation of images from an object placed at a desired distance from the lens.Refraction of light takes place when it is travelling from one medium to another, that is, when there is a difference in the medium it is travelling in. Ray diagrams are pictorial representations of how the light from an object placed near a lens will refract when passing through the lens and form a corresponding image of the object depending upon where the object was placed.

Thus, the distance of the object from the principal focus, that is the center point of the lens, is known as the object distance denoted by $u$ while the distance from the principal focus to the image that is formed is called the image distance denoted by $v$. The focal length is the distance of the focus from the lens’ optical center. Unlike mirrors, concave and convex lenses are capable of producing both virtual and erect images as well as real and inverted images depending upon the location or the position at which the object is placed.

Even-though convex and concave lenses have differing properties the formula that relates the focal length, image distance and object distance known as the thin lens formula is the same for both of these types of lenses. This thin lens formula was derived considering certain assumptions. The basic assumptions that were taken include; the use of thin lenses with small aperture for which the object considered was also taken to be small. The angle made by the incident and reflected rays were also considered to make small angles with the principle axis of the thin lens. The thin lens equation is hence given as:
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ ---------($1$)

Let us now extract the data given in the question. The focal length is given and the distance at which the image is formed is also given to be $18\;cm$ and $24\;cm$ respectively. The image is said to be formed on the other side of the lens. We are asked the distance at which the object has been placed which means we are required to find the object distance.

Given, $f = 18\;cm$ and $v = 24\;cm$.By substituting these above given values into equation ($1$) we get:
$\dfrac{1}{{18}} = \dfrac{1}{{24}} - \dfrac{1}{u}$
By rearranging the terms we get:
$\dfrac{1}{u} = \dfrac{1}{{24}} - \dfrac{1}{{18}}$
Taking the LCM we get:
$\dfrac{1}{u} = \dfrac{{3 - 4}}{{72}}$
We now solve the above equation to get:
$\dfrac{1}{u} = - \dfrac{1}{{72}}$
By cross-multiplying the terms we get:
$72 = - u$
By rearranging we get:
$u = - 72\;cm$

Hence, this is the object distance, that is, the distance at which the object should be placed to get an image to be formed at $24\;cm$. Next, we are asked to find the magnification. The magnification refers to how the image size has varied from the size of the object and this change is said to be linear with respect to the object size hence the magnification produced by the thin lens is called linear magnification. Linear magnification is defined as the ratio of the size of the image produced by the lens to the size of the object. The formula for magnification in terms of the image and object distance is given as:
$m = \dfrac{v}{u}$ ----------($2$)

To determine the magnification of the image, the object distance which was found out and the given image distance values must be substituted in equation ($2$). Thus, we get:
$m = \dfrac{{24}}{{ - 72}}$
$ \Rightarrow m = - \dfrac{{24}}{{72}}$
By reducing this into simpler fraction we get:
$m = - \dfrac{1}{3}$
Hence, this is the magnification that is produced by the thin lens. The magnification also gives the nature of the image as well. The negative sign indicates that the image formed is real and inverted in accordance to the data given in the question that the image is formed on the other side of the lens. Also, since the magnitude of $m$ is less than one which specifies that the image formed is diminished.

Additional information: Concave lenses are observed to produce diverging properties which means that when light rays are incident on the surface of the concave lenses then they tend to diverge, that is reflect away from the lens unlike convex lenses that tend to converge the light rays. The divergent light rays in this case of concave lenses and the converged rays in the case of convex lenses meet at a point in order to form the image of the object that is considered.

Note: Convex lenses are lenses that are bulged at the center and sleek at the ends of the spherical lens while concave lenses are ones that are thin at the center and bulged at the two ends of the lens. The common misconception may occur in the application of sign conventions which is not required here as the focal length and image distance values are already given in the question after applying the conventions.