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Question: An object is placed at a distance of \[40\] cm in front of a concave mirror of focal length \[20\] c...

An object is placed at a distance of 4040 cm in front of a concave mirror of focal length 2020 cm. Determine the ratio of the size of the image and the size of the object.
A. 2:12:1
B. 1:21:2
C. 1:11:1
D. 4:14:1

Explanation

Solution

Given both the object distance uu and the focal length ff of the concave mirror. To find the ratio of the size of the image to the size of the object we need to find the magnification of the given concave mirror. The ratio of the size of the image to the size of the object of the spherical mirror is known as the magnification of the spherical mirror. For that, we need to find the image distancevv using the mirror formula.

Complete step by step solution:
The object distance is given as u=−40cmu = - 40cm
Given that focal length of the concave mirror is given as f=−20cmf = - 20cm
Therefore by the mirror formula,
1f=1u+1v\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}
1v=1f−1u\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}
Substituting the corresponding values,
1v=1−20−1−40\dfrac{1}{v} = \dfrac{1}{{ - 20}} - \dfrac{1}{{ - 40}}
⇒1v=1−20+140\Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 20}} + \dfrac{1}{{40}}
On simplification
1v=−2+140\dfrac{1}{v} = \dfrac{{ - 2 + 1}}{{40}}
⇒1v=1−40\Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 40}}
Therefore the image distance v=−40cmv = - 40cm
Now using the magnification formula,
m=vum = \dfrac{v}{u}
Therefore substituting the values,
m=(−40−40)m = (\dfrac{{ - 40}}{{ - 40}})
Therefore the ratio is 1:11:1
Therefore the correct option is C.

Note:
Ray diagrams require sign convention to make sense. These sign conventions are required to measure distances in the case of spherical mirrors. In spherical mirrors, the right is considered as the polished surface. So the reflecting surface and the object will be on the left side. The pole of the mirror will be placed on the origin. This will be similar to the Cartesian coordinate as we consider everything on the right side as positive and the left side as negative. Similarly, everything that is on the right side of the pole will be positive and the left side will be negative. And that is why we have taken on the problem everything as negative because in the question it is said that an object is placed in front of the concave mirror.