Question

Find the nature of image when an object is placed at 2f from the pole of a convex mirror of focal length f
A) Real, inverted and of same size.
B) virtual, erect and of $\dfrac{1}{3}$ size.
C) virtual, erect and of 3 times.
D) real, inverted and $\dfrac{1}{3}$times.

Answer 1

Hint
Use the mirror formula to determine the position of the final image and after that calculate the magnification and interpret the answer. The note at the end of the answer will help in interpreting the correct answer by eliminating options.

Complete step by step answer
Given, object is placed at 2f from the pole
$\therefore$ $u = - 2f$(using the sign convention )
Now, using the mirror formula for this object we have,
$\Rightarrow \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Now, putting the values given in the question we have,
$\Rightarrow \dfrac{1}{v} + \dfrac{1}{{ - 2f}} = \dfrac{1}{f}$
On further solving we have,
$\Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{{2f}} $
$\Rightarrow \dfrac{1}{v} = \dfrac{3}{{2f}} $
$\Rightarrow v = \dfrac{{2f}}{3}$
Positive sign of v tells us that the image is formed behind the mirror. Hence the image must be virtual.
Now, magnification is given by,
$\Rightarrow m = \dfrac{{ - v}}{u}$
Putting the values we have,
$\Rightarrow m = \dfrac{{ - \dfrac{{2f}}{3}}}{{ - 2f}} $
$\Rightarrow m = \dfrac{1}{3} $
Positive sign of magnification tells us that the image is erect and the value of magnification i.e. $\dfrac{1}{3}$ implies that the height of image is $\dfrac{1}{3}$ of the object height.
Hence taking all this into account we can conclude by saying that the correct answer is option (B).

Note
i) Positive sign of magnification tells us that the image is erect and negative sign of magnification tells us that the image is inverted.
ii) Positive sign of v tells us that the image is formed behind the mirror and negative sign of v tells us that the image is formed in front of the mirror.