Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of the image.

Answer 1

We need to understand the relation between the size of an object, its distance from mirror and the focal length of the concave mirror with the image formed by the mirror and the properties of the image such as the size, position and nature.

Complete answer:
We are given a concave mirror which has a focal length of 18 cm. An object of height 7 cm is placed in front of the concave mirror such that it is 27 cm away from the mirror.
We know that the image formed by this object, if real can be captured on a screen. The position of the screen with respect to the mirror can be calculated using the mirror formula which relates the focal length ‘f’ of the mirror, the object distance ‘u’ and the image distance ‘v’. It is given as –

& \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \\\ & \Rightarrow \dfrac{1}{-18cm}=\dfrac{1}{v}+\dfrac{1}{-27cm} \\\ & \Rightarrow \dfrac{1}{v}=\dfrac{1}{27}-\dfrac{1}{18} \\\ & \Rightarrow \dfrac{1}{v}=\dfrac{2-3}{54} \\\ & \therefore v=-54cm \\\ \end{aligned}$$  The image of the object which was kept 27 cm away from the mirror is formed at 54 cm in front of the mirror. The screen can be placed 54 cm to the left of the mirror to get the image. Now, we can find the size of the image formed using the ratio relationship between the image distance and object distance to the height of the image and the object as – $$\begin{aligned} & \dfrac{h'}{h}=-\dfrac{v}{u} \\\ & \Rightarrow \dfrac{h'}{7cm}=-\dfrac{-54}{-27} \\\ & \Rightarrow h'=-7cm\times 2 \\\ & \therefore h'=-14cm \\\ \end{aligned}$$ We get the image size to be double the object size with a negative sign. This means that the image formed will be real and inverted in nature. The image formed by the object in front of a concave mirror is 14 cm in size, is 54 cm in front of the mirror and is real and inverted in nature. This is the required solution. **Note:** The signs involved in calculating the image distance from the mirror and the magnification or the size of the image should be treated very carefully. Any change in the sign can result in totally different solution which will not coincide with the observation