Question

A concave mirror produces three times magnified (enlarged) real images of an object placed at 10 cm in front of it. Where is the image located?

Answer 1

The magnification is image distance by the object distance. Using this formula we will find the image distance. The mirror formula relates the terms, such as, the image distance, the object distance and the focal length. Using this formula we will find the value of the focal length.
Formula used:
$m=-\dfrac{v}{u}$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

Complete answer:
From the given information, we have the data as follows.
A concave mirror produces three times magnified (enlarged) real images of an object placed at 10 cm in front of it.
The magnification formula in terms of the image distance and the object distance is,
$m=-\dfrac{v}{u}$
Substitute the values in the above formula.
$3=-\dfrac{v}{-10}$
Continue further computation.

& v=-(3\times -10) \\\ & \therefore v=30\,cm \\\ \end{aligned}$$ The distance between the mirror and image is 30 cm. **Additional information:** The mirror formula gives the relation between the image distance, object distance and the focal length. $$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$ Substitute the values in the above formula. $$\dfrac{1}{30}+\dfrac{1}{-10}=\dfrac{1}{f}$$ Continue further computation. $$\begin{aligned} & \dfrac{1}{f}=\dfrac{1}{30}-\dfrac{1}{10} \\\ & \therefore f=-15 \\\ \end{aligned}$$ The focal length of the lens is, - 15 cm. The radius of curvature in terms of the focal length is, $$R=2f$$ Substitute the values in the above formula. $$R=2\times -15$$ Continue further computation. $$\therefore R=-30\,cm$$ The radius of curvature of the lens is, - 30 cm. $$\therefore $$The image is located 30 cm in front of the mirror. **Note:** Using the relation between the focal length and the radius of curvature, we will find the value of the radius of curvature. The mirror formula relates the terms, such as, the image distance, the object distance and the focal length.