Question

A small candle \[{\mathbf{2}}.{\mathbf{5}}{\text{ }}{\mathbf{cm}}\] is size is placed at \[{\mathbf{27}}{\text{ }}{\mathbf{cm}}\] in front of a concave mirror of radius of curvature \[{\mathbf{36}}{\text{ }}{\mathbf{cm}}\]. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of image. If the candle is moved closer to the mirror how would the screen have to be moved?

Answer 1

Hint: Use the mirror formula, \[\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\] and find the image distance. Use the formula \[m = - \dfrac{v}{u} = \dfrac{{h'}}{h}\] and find the image height.

Complete step by step solution:
In this given problem,

Distance of the object, \[u = - 27\,{\text{cm}}\]
Height of the object, \[h = 2.5\,{\text{cm}}\]
Radius of curvature, \[R = - 36\,{\text{cm}}\]
Focus,
 \[
  f = - \dfrac{{36}}{2}\,{\text{cm}} \\
  {\text{ = }} - 18\,{\text{cm}} \\
\]
(The negative sign is due to the type of the mirror; it is negative in case of a concave mirror as it is a diverging mirror.)
Let the image distance and height of the image be \[v\] and \[h'\].

The mirror formula is given by:
\[\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}\] …… (1)

Now, substitute \[u = - 27\,{\text{cm}}\] and \[f = - 18\,{\text{cm}}\]in the equation (1)
\[
  \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f} \\
   - \dfrac{1}{{27}} + \dfrac{1}{v} = - \dfrac{1}{{18}} \\
  \dfrac{1}{v} = \dfrac{1}{{27}} - \dfrac{1}{{18}} \\
  \dfrac{1}{v} = \dfrac{{2 - 3}}{{54}} \\
  \dfrac{1}{v} = \dfrac{{ - 1}}{{54}} \\
  v = - 54\,{\text{cm}} \\
\]

Hence, the image distance is \[ - 54\,{\text{cm}}\].

The magnification formula is:
\[m = - \dfrac{v}{u}\] …… (2)
Magnification formula can also be written as:
\[m = \dfrac{{h'}}{h}\] …… (3)

Comparing the equations (2) and (3), we get,
\[ - \dfrac{v}{u} = \dfrac{{h'}}{h}\] …… (4)
Substituting, the values of \[u = - 27\,{\text{cm}}\], \[h = 2.5\,{\text{cm}}\]and \[v = - 54\,{\text{cm}}\]in equation (4), we get,
\[
   - \dfrac{{ - 54}}{{ - 27}} = \dfrac{{h'}}{{2.5}} \\
  h' = \dfrac{{ - 54 \times 2.5}}{{27}} \\
  h' = - 5\,{\text{cm}} \\
\]

The height of the image is \[ - 5\,{\text{cm}}\].

(Negative sign indicates that the image is inverted.)
The nature of the image is virtual and inverted as the image is formed behind the mirror.

If the candle is placed closer to the mirror, then the screen must be placed away from the mirror to get the image.

Note: In this problem, you are asked to find the height of the image and its nature. Always take care of the object distance as it is always negative according to sign convention, while not doing so will definitely affect the result. Height of the image may come out to be negative, that is nothing to worry about. Negative value indicates that the image is inverted.