A convex mirror of focal length f forms an image \( \frac{1}{n} \) times the size of the object. The distance of the object from the mirror is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (n-1)f \)
A convex mirror always forms a virtual, erect and diminished image. So the magnification \( m=-v/u \) is positive, that is, \( m=+1/n \). With the Cartesian sign convention, the object distance \( u \) is negative. The focal length \( f \) of a convex mirror is positive in the mirror equation \( 1/v+1/u=1/f \).
From the magnification:
\( \displaystyle -\frac{v}{u}=\frac{1}{n} \)
\( \displaystyle \Rightarrow\; v=-\frac{u}{n} \)
Putting this in the mirror equation:
\( \begin{aligned} \frac{1}{f} &= \frac{1}{v}+\frac{1}{u} \\ &= -\frac{n}{u}+\frac{1}{u} \\ &= \frac{1-n}{u} \end{aligned} \)
\( \begin{aligned} u &= (1-n)\,f \\ &= -(n-1)\,f \end{aligned} \)
The negative sign shows that the object is in front of the mirror. Therefore, the distance of the object from the mirror is \( (n-1)f \).