Question

An object 1 cm high produces a real image 1.5 cm high, when placed at a distance of 15 cm from a concave mirror. The position of the image is

Answer 1

Hint: We need to apply the mirror equation to calculate the focal length. Then the position is calculated by further evaluations.
Formula used: To calculate the position, we use the mirror equation:
$\dfrac{{\text{1}}}{{\text{f}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{u}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{v}}}$
Here, ${\text{f}}$ is the focal length of the mirror.
${\text{u}}$ is the distance of the object from the mirror.
${\text{v}}$ is the distance of the image from the mirror.

Complete step by step answer:
We need to calculate the position of the image and the magnification,
As we already know:
${\text{m = }}\dfrac{{{\text{ - v}}}}{{\text{u}}}{\text{ = }}\dfrac{{{{\text{h}}_{\text{i}}}}}{{{{\text{h}}_{\text{o}}}}}$
$ \Rightarrow \dfrac{{{\text{ - v}}}}{{{\text{ - 15}}}}{\text{ = }}\dfrac{{{\text{ - 1}}{\text{.5}}}}{{\text{1}}}$
By the above formula we easily get the image distance.
${{v = - 15 \times 1}}{\text{.5 = - 22}}{\text{.5}}$
Hence, the magnification is:
${\text{m = }}\dfrac{{{\text{ - 22}}{\text{.5}}}}{{{\text{ - 15}}}}{\text{ = - 1}}{\text{.5}}$
So, the position of the image 22.5 behind the mirror.

Additional Information : Spherical mirrors are the mirrors having curved surfaces that are painted on one of the sides. Spherical mirrors in which inward surfaces are painted are known as convex mirrors, while the spherical mirrors in which outward surfaces are painted are known as concave mirrors. If a hollow sphere is cut into parts and the outer surface of the cut part is painted, then it becomes a mirror with its inner surface as the reflecting surface. This type of mirror is known as a concave mirror.

Note: According to the sign convention of the mirror: Object is always placed to the left of the mirror. All distances are measured from the pole of the mirror. Distances measured in the direction of the incident ray are positive and the distances measured in the direction opposite to that of the incident rays are negative. It should be always taken care before solving the question.