Question

A lamp is 5 m from a wall. Find the focal length of a concave mirror which will form a four times magnified image of the lamp on the wall.
(A) 4 m
(B) 2 m
(C) 20 m
(D) 10 m

Answer 1

Hint:While solving all these kinds of problems we need to be very careful about the sign conventions. All the distances to the left of the pole of the mirror are to be measured in negative and all to the right of the pole in positive. Similarly, all the distances above the principal axis are to be measured in positive while all below the principal axis to be measured in negative. Also, for convex mirrors the focal length is always positive while for the conv mirror the focal length is always negative.

Complete step by step answer:
Given Concave mirror
Object distance, u= -5 m
Given that the image is four times magnified and we know concave mirror always forms magnified image when the image is real, so $m=-4$
Now the relationship between the magnification and other quantities for a mirror is given as: $m=\dfrac{-v}{u}$, substituting the values,
$\Rightarrow -4=\dfrac{-v}{-5}$
$\Rightarrow v=-20$cm
Now we need to find the focal length, we make use of mirror formula:\[\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
\[\Rightarrow \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}\]
\[\Rightarrow \dfrac{1}{f}=\dfrac{-1}{20}-\dfrac{1}{5}\]
\[\Rightarrow \dfrac{1}{f}=\dfrac{-1-4}{20}\]
\[\therefore f=-4m\]

So, the correct option is A.

Note:We need to be very careful while solving such problems as we have to use the sign conventions. Also, there are some tricks to cross verify our solution. We need to keep in mind that the real image is always inverted while the virtual image is always erect. The sign of magnification also reveals something. If magnification is positive then image is erect and if magnification is negative then image is inverted.