Question

The image produced by a concave mirror is one-quarter the size of the object. If the object is moved 5 cm closer to the mirror, the image will only be half the size of the object. The focal length of mirror is:
 (A) $f=5.0cm$
 (B) $f=2.5cm$
 (C) $f=7.5cm$
 (D) $f=10cm$

Answer 1

Hint It should be known to us that for the thin lens in air, the focal length is the distance from the centre of the lens to the principal foci, or we can say the focal points of the lens. For the converging lens, for example in case of the concave lens, the focal length is positive and is the distance at which a beam of collimated light will be focused on a specific point. Using this concept we can solve this question.

Complete step by step answer
Let us consider that the initial distance of the object is u.
The magnification is $\dfrac{1}{4}$
So, we can say that $v=\dfrac{u}{4}$
Hence, we can write that:
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$
$\Rightarrow \dfrac{-4}{u}+\dfrac{-1}{u}=\dfrac{1}{f}.................(1)$
Now the object moves 5 cm closer to the mirror.
So,
${{u}^{/}}=-(u-5)$and magnification is $\dfrac{1}{2}$,
So, ${{v}^{/}}=-\dfrac{(u-5)}{2}$
Hence, we can write that:
$\dfrac{1}{{{v}^{/}}}+\dfrac{1}{{{u}^{/}}}=\dfrac{1}{f}$
$\Rightarrow \dfrac{-2}{u-5}+\dfrac{-1}{u-5}=\dfrac{1}{f}............(2)$
From the equation (1) and (2) we get that:
$\dfrac{4}{u}+\dfrac{1}{u}=\dfrac{2}{u-5}+\dfrac{1}{u-5}$
From this expression we get that the value of u is 12.5.
Now we have to substitute the value u in the equation 1 to get:
$\dfrac{-4}{12.5}+\dfrac{-1}{12.5}=\dfrac{1}{f}$
The value of f is -2.5. Here the negative sign indicates that it is a concave mirror.

Hence the correct answer is option B.

Note It should be known to us that concave mirrors can produce both the real and virtual images. The images can be upright in case they are virtual and inverted if they are real. The images are behind the mirror if they are virtual or in the front of the mirror if they are real. The images can also be enlarged, reduced or of the same size as the object.