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# The focal length of a concave mirror is \$12cm\$. Where should an object of length \$4cm\$ be placed so that a real image of length \$1cm\$ is formed?A) \$48cm\$B) \$3cm\$C) \$60cm\$D) \$15cm\$

Last updated date: 16th May 2024
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Hint:A mirror concave or convex has a definite radius of curvature, focal length and the images produced by them vary differently in different scenarios. We can easily determine the position, height of the object if these respective values are known for the image formed by the mirror by using the mirror formula.

Formula used:
Mirror formula
\$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\$
Where \$u\$ is the distance of object from the pole of the mirror
\$v\$ is the image distance from the pole
\$f\$ is the focal length of the mirror

Complete step by step answer:
In a spherical mirror when using the mirror formula we use certain sign conventions. Let’s discuss them first.
Anything measured towards right and up of the pole of the spherical mirror is taken as positive and anything measured towards the left or down of the pole of the mirror is considered negative. Just like the two-dimensional quadrant system.
So the object distance for an erect object in case of a concave mirror is always negative whereas its height is positive.
Let’s now understand what a real-image is. When the image formed by a mirror can be captured by screen is called real image. For an image being capable of capturing on a screen it must get projected by the mirror towards the object side.
Thus for this case we can say that the image distance is negative as per sign convention.
Now what about the focal length of the concave mirror? Well as we know, the concave mirror diversifies the light rays falling on its reflecting surface. This means that the focal length of the concave mirror is also assigned a negative sign.
Our above analysis gives us the mirror formula for the concave mirror.
Mathematically,
\[(\dfrac{1}{{ - u}}) + (\dfrac{1}{{ - v}}) = (\dfrac{1}{{ - f}})\]
Or simply speaking,
\$ \Rightarrow \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\$
Magnification of the image is given by the following formula in spherical mirrors.
\$m = - \dfrac{{h'}}{h}\$ ……. (1)
Where \$h'\$ is the height of the image and \$h\$ is the height of the object.
In the above equation the negative sign implies that when a real image is formed by a concave mirror it is inverted.
Otherwise the magnification formula can also be expressed in terms of focal length and object distance as follows
\$m = \dfrac{f}{{f - u}}\$
This formula as per sign convention becomes
\$m = \dfrac{{ - f}}{{u - f}}\$ ……… (2)
Comparing equation (1) and (2) we get
\$ \Rightarrow \dfrac{{ - h'}}{h} = \dfrac{{ - f}}{{ - f + u}}\$
Substituting the respective values in the above equation we get,
\$ \Rightarrow \dfrac{{ - 1}}{4} = \dfrac{{ - 12}}{{ - 12 + u}}\$
\$ \Rightarrow 12 - u = - 48\$
Simplifying we get
\$u = - 60cm\$

Thus our correct answer is option (C).

Note:In any problem of optics always use proper sign convention. And also be careful to convert the units to a single system of units before attempting calculations to avoid errors as well as saving time. We have elaborated the mirror formula for the real image of a concave mirror in order to make easy understanding of sign conventions used in mirror equations.