Question

An object is placed \[{\text{15 cm}}\] from a convex mirror of curvature \[{\text{90 cm}}\].Calculate the image position and the magnification.

Answer 1

Hint: In this question, we have been asked to find out the image position and magnification of the convex mirror. As we know, the radius of curvature of a mirror is twice the focal length. We can use this formula to find out the focal length. Then use the mirror formula to find the image position and the magnification formula of a mirror to find out the magnification.

Formula used:
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
where, \[v = \] image distance, \[u = \] object distance, \[f = \]focal length, \[f = \dfrac{R}{2}\] and \[R = \] Radius of curvature

Complete step by step answer:
Let us write down the given data, Object distance from convex mirror, \[u = - 15{\text{ cm}}\].The Negative sign indicates object is in the opposite direction of the incident light.The Radius of curvature of the convex mirror, \[{\text{R = 90 cm}}\].Focal length of the convex mirror,
\[f = \dfrac{R}{2} \\
\Rightarrow \dfrac{{90}}{2} = 45{\text{ cm}}\]
Focal length is positive since it is in the direction of incident light

We need to find image distance from the convex mirror, \[v\]. Using mirror formula to find \[v\] we get,
\[\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}\]
Substituting the values of \[u\] and \[f\] respectively we get,
\[ \Rightarrow \dfrac{1}{v} + \left( { - \dfrac{1}{{15}}} \right) = \dfrac{1}{{45}}\]
\[ \Rightarrow \dfrac{1}{v} - \dfrac{1}{{15}} = \dfrac{1}{{45}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{45}} + \dfrac{1}{{15}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{{1 + 3}}{{45}}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{4}{{45}}\]
Thus, from this we get,
\[ \Rightarrow v = \dfrac{{45}}{4}\]
\[ \Rightarrow v = 11.25{\text{ cm}}\]

Therefore, the image position is at \[11.25{\text{ cm}}\] behind the mirror.Magnification of the mirror is given by,
\[m = - \dfrac{v}{u}\]
On substituting the values of \[v\] with \[11.25{\text{ cm}}\] and \[u\] with \[ - 15{\text{ cm}}\] in the formula we get.
\[m = - \dfrac{{\left( {11.25} \right)}}{{\left( { - 15} \right)}} \\
\Rightarrow m= \dfrac{{11.25}}{{15}}\]
\[ \therefore m = + 0.75\]

Therefore, magnification is \[ + 0.75\] implies, the image is erect, virtual, and diminished.

Note: Here we need to be careful about the signs if the object is placed in the opposite direction of the incident light, we need to put a negative sign in the object’s distance and if the object is placed in the same direction of the incident light, we put a positive sign. For convex mirror focal length is always positive as it is in the same direction as incident light.