Question

A real object is placed at a distance $f$ from the pole of a convex mirror, in front of the convex mirror. If the focal length of the mirror is $f$, then distance of the image from the pole of the mirror is:
A. $2f$
B. $\dfrac{f}{2}$
C. $4f$
D. $\dfrac{f}{4}$

Answer 1

To solve this question, we will use the mirror formula that gives a relation between the focal length of the mirror, the distance of the object from the pole of the mirror and the distance of the image from the pole of the mirror. Also, we need to take care about the negative or positive signs that should be used while using this formula.
Formula used:
$\dfrac{1}{F}=\dfrac{1}{v}+\dfrac{1}{u}$

Complete step by step answer:
The mirror formula is as follows:
$\dfrac{1}{F}=\dfrac{1}{v}+\dfrac{1}{u}$
Here, $F$ is the focal length of the mirror used,
$v$ is the distance of the image from the optical center of the mirror,
$u$ is the distance of the object from the optical center of the mirror.
Now, let us consider the given question, it is given in the question that the focal length of the convex mirror is $f$, the distance of the object from the optical center of the mirror i.e., from the pole of the convex mirror is given as $f$. Since the object is placed in front of the convex mirror hence a negative sign sill be added to the distance, thus $u=-f$. Now, let us substitute all three values in the above stated formula:

& \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{-f} \\\ & \Rightarrow \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{f} \\\ & \Rightarrow \dfrac{1}{v}=\dfrac{2}{f} \\\ & \therefore v=\dfrac{f}{2} \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** The mirror formula should be used very carefully, keeping in mind the correct sign conventions. Distances are positive if they are measured along the direction of the incident ray. If the incident ray is travelling in the opposite direction, then the distance should be negative.