An object placed \[10\;{\rm{cm}}\] in front of a lens has an image \[20\;{\rm{cm}}\] behind the lens. What is the power of lens (in dioptres)
A. 1.5
B. 3.0
C. -15.0
D. +15.0

Hint: The above problem can be resolved using the concepts and fundamentals of the power formula of the lens. In this case, we are provided with the information regarding the distance of the object and distance of image formed. Using this information and the mathematical expression for the lens formula, we can find the value of the focal length of the lens. This value of focal length is utilised to calculate the power of the lens by making suitable substitution of values.

Complete step by step answer:
Given:
The object distance is, \[u = 10\;{\rm{cm}}\].
The image distance is, \[v = 20\;{\rm{cm}}\].
Apply the lens formula to calculate the focal length as,
\[\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
As, the object is placed in front of the lens and image is formed behind the lens. Thus, the sign convention for the u is negative and that of v is positive.
Solve by substituting the values in above equation as,
 \[
\Rightarrow \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{20\;{\rm{cm}}}} - \dfrac{1}{{\left( { - 10\;{\rm{cm}}} \right)}}\\
\Rightarrow \dfrac{1}{f} = \dfrac{1}{{20\;{\rm{cm}}}} + \dfrac{1}{{10\;{\rm{cm}}}}\\
\Rightarrow f = \dfrac{{20}}{3}\;{\rm{cm}}.........................................\left( 1 \right)
\]
 Then apply the formula for the power of lens as,
\[\Rightarrow P = \dfrac{{100}}{f}\]
Solve by substituting the value of equation 1 in above formula as,
\[
\Rightarrow P = \dfrac{{100}}{f}\\
\Rightarrow P = \dfrac{{100}}{{\left( {\dfrac{{20}}{3}\;{\rm{cm}}} \right)}}\\
\Rightarrow P = 15\;{\rm{D}}
\]
Therefore, the power of the lens is 15 Dioptres and option (D) is correct.

Note:To resolve the given problem, one must understand the concept and applications of the lens formula. The lens formula is utilised for calculating the variables like image distance., object distance and the focal length. These variables possess some sign conventions, which is decided accordingly with the position of the object and the position of the image formed. Moreover, it is also required to remember this formula to solve many such cases.