Question

A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image. Describe the nature of size of the image. If the candle is moved closer to the mirror, how should the screen have to be moved.

Answer 1

The image obtained by the concave mirror is generally real and inverted. The focal length of the mirror is half of the radius of the curvature of the mirror. The incident ray when passes through the point of centre of curvature represented by R is not deviated.

Formula used: The mirror formula is given by,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Where focal length is f the image distance from the centre of the mirror is v and the object distance from the centre of the mirror is u.
The formula of the magnification is given by,
$\Rightarrow m = \dfrac{{h'}}{h} = - \dfrac{v}{u}$
Where the magnification is m the image height is $h'$ and the object height is$h$. The image distance is v and the object distance is u.

Complete step by step solution:
It is given in the problem that a small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm and we need to tell the object distance that need place in order to get a sharp image also how we need to find how to move the screen.
The focal length of the mirror is given by,
$\Rightarrow f = \dfrac{R}{2}$
The focal length of the mirror is equal to,
$\Rightarrow f = \dfrac{R}{2}$
$\Rightarrow f = - \dfrac{{36}}{2}$
$\Rightarrow f = - 18cm$.
The mirror formula is given by,
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Where focal length is f the image distance from the centre of the mirror is v and the object distance from the centre of the mirror is u.
The object distance is equal to $u = - 27cm$ the radius of curvature is$R = - 36cm$
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{{ - 18}} = \dfrac{1}{v} + \dfrac{1}{{ - 27}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{{27}} - \dfrac{1}{{18}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{2 - 3}}{{54}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{54}}$
$\Rightarrow v = - 54cm$.
The image distance is equal to$v = - 54cm$.
The formula of the magnification is given by,
$\Rightarrow m = \dfrac{{h'}}{h} = - \dfrac{v}{u}$
Where the magnification is m the image height is $h'$ and the object height is$h$. The image distance is v and the object distance is u.
The image distance is $v = - 54cm$ and the object distance is equal to$u = - 27cm$, the magnification is equal to,
$\Rightarrow m = - \dfrac{v}{u}$
$\Rightarrow m = - \dfrac{{\left( { - 54} \right)}}{{\left( { - 27} \right)}}$
$\Rightarrow m = - \dfrac{{54}}{{27}}$
$\Rightarrow m = - 2$
The height of the image is equal to,
$\Rightarrow m = \dfrac{{h'}}{h}$
$\Rightarrow - 2 = \dfrac{{h'}}{{2 \cdot 5}}$
$\Rightarrow h' = - 2 \cdot 5 \times 2$
$\Rightarrow h' = - 5cm$
The image height is 5 cm and the image is virtual and inverted.
The candle is moved towards the mirror therefore the screen has to move further away from the candle in order to obtain a sharp image.

Note: The object distance is taken as negative if the object is placed on the left side of the mirror and the image distance of the object will be positive or negative depending on whether the image is formed on the right hand side or the left hand side of the mirror. The focal length of the mirror is negative as in the case of the concave mirror the focus is on the left hand side of the mirror.