Question

An object 5.0cm in length is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature, and its size.

Answer 1

Hint: The convex and concave mirror are the two sides of a circular portion of a sphere. If you consider a circular portion of a sphere, applying the reflective coating on the outer side will make it a concave mirror and applying the reflective coating on the inner side will make it a convex mirror.
Irrespective of concave or convex mirror, the image formed will follow the mirror equation,
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
where, f = focal length of the mirror, v = distance of image from pole, u = distance of object from pole.

Complete step by step solution:
Consider the object placed at a distance of 20cm in front of a convex mirror of radius of curvature, R = 30cm.
Distance of the object, u = – 20 cm
Focal length, $f = \dfrac{R}{2} = \dfrac{{30}}{2} = 15cm$
There are two rays emanating from the object,
i) First ray passes parallel to the principal axis and after deflection, when traced back, passes through primary focus, f.
ii) Second ray passes along the direction of secondary focus, 2f and retraces the same path as shown.
The image is formed at distance +v from the pole.
Applying the mirror equation, we get –
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{15}} - \dfrac{1}{{\left( { - 20} \right)}}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{15}} + \dfrac{1}{{20}} = \dfrac{7}{{60}}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{15}} + \dfrac{1}{{20}} = \dfrac{7}{{60}}$
$\therefore v = \dfrac{{60}}{7} = 8.57cm$
The image is formed at a distance of 8.57 metres from the convex mirror.
Magnification is defined as the ratio between the height of the object to the height of the image. It represents the change in the height of the object after the reflection.
The formula for magnification in a mirror is –
$m = - \dfrac{v}{u} = - \dfrac{{8.57}}{{\left( { - 20} \right)}} = \dfrac{{8.57}}{{20}} = 0.428$
Since, the magnification is positive, the image is virtual and since it is lesser than one, it is diminished.

Therefore, The image is formed at a distance of 8.57 m from the pole and it is virtual and diminished.

Note: The students should exercise extreme caution while substituting the values of u and v in the equations. As you can see, in the above steps, the value of u is substituted with entire brackets first, and then, solved by removing the brackets. This is the best practice to avoid errors due to sign convention in these problems.