Question

An object 4cm in size is placed 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should the screen be kept so that a sharp image of the object can be obtained on screen? Find the nature and height of the image.

Answer 1

Hint: To solve this question, we will use the mirror formula, which has three values in it, that are focal length of the mirror(f), distance of object from the mirror(u) and distance of image from the mirror(v). If two of them are known, the other third value can be calculated using the formula: \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\].
Formula used: \[\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}\], $magnification = \dfrac{{{h_2}}}{{{h_1}}} = - \dfrac{v}{u}$.

Complete Step-by-Step solution:
A formula which gives the relationship between image distance(v), object distance(u) and focal length(f) of the spherical mirror is known as the mirror formula. The mirror formula can be written as:
$ \Rightarrow \dfrac{1}{{{\text{image distance}}}} + \dfrac{1}{{{\text{object distance}}}} = \dfrac{1}{{{\text{focal length}}}}.$
$ \Rightarrow \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. ………………..(i)
Where u = distance of object from the mirror.
              v = distance of image from the mirror.
              f = focal length of the mirror.
Given that, size of object, ${h_1} = 4cm$, distance of object from the mirror, u = 25cm, focal length of the mirror, f = 15cm.
We have to find out at what distance the screen should be placed so that a sharp image of the object can be obtained.
We know that the object distance, u is always negative and here, the mirror is a concave mirror. So, the focal length is also considered as negative.
Now from equation (i), we get
$ \Rightarrow \dfrac{1}{v} + \dfrac{1}{{( - u)}} = \dfrac{1}{{( - f)}}$.
$
   \Rightarrow \dfrac{1}{v} - \dfrac{1}{u} = - \dfrac{1}{f} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{u} - \dfrac{1}{f} \\
$
Putting the values of u and f, we get
$
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{{25}} - \dfrac{1}{{15}} = \dfrac{{3 - 5}}{{75}} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{ - 2}}{{75}} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 37.5}} \\
   \Rightarrow v = - 37.5cm \\
$
Here we get a negative value of v. this means that the image is formed in front of the mirror (left side) at a distance of 37.5cm from the pole of the concave mirror.
The nature of the image is real and inverted and the image is formed below the principal axis.
We know that, $magnification = \dfrac{{{h_2}}}{{{h_1}}}$ ………(ii)
where ${h_2}$is the height of the image and ${h_1}$ is the height of the object.
The magnification can also be identified by, $magnification = - \dfrac{v}{u}$. …………(iii)
Comparing equation (ii) and (iii), we get
$ \Rightarrow \dfrac{{{h_2}}}{{{h_1}}} = - \dfrac{v}{u}$
We have u = -35cm, v = - 37.5cm and ${h_1}$= 4cm.
$
   \Rightarrow \dfrac{{{h_2}}}{{4cm}} = - \dfrac{{( - 37.5)cm}}{{( - 25)cm}} \\
   \Rightarrow \dfrac{{{h_2}}}{{4cm}} = - \dfrac{{37.5}}{{25}} \\
   \Rightarrow {h_2} = 4 \times ( - 1.5)cm \\
   \Rightarrow {h_2} = - 6cm \\
$
Height of all the real and inverted images is always negative.
Hence, we can say that the height of the image formed is 4cm.
Therefore, the nature of the image is real and inverted and the height of the image is 4 cm.

Note: Whenever we ask such types of questions, we have to remember some basic points of the mirror formula. We should know all the sign conventions for spherical mirrors given in the question then we will put the given values with their appropriate signs in the mirror formula. By solving the formula step by step, we will get the required values. After that we will use the formula of linear magnification to find the unknown values by putting the known ones.