Question

A real object is placed perpendicular to the principal axis of a concave mirror at a position 1 such that the image formed is real with a magnification 2. The object is now shifted to another position 2 at a distance $!5cm$ from position 1 and a real image is obtained with a magnification $8$. Shifting the position of object from 2 to a third position 3 gives an image with magnification $1/2$
Focal length of the mirror is

• A. $-6cm$
• B. $-4cm$
• C. $-3cm$
• D. none of these

Answer 1

Answer: (D)

none of these

$\begin{array}{c}\text{Real and magnified Image is formed}\\ \text{when object is between}c\text{and}f\\ \text{magnification increases on moving from}\\ c\text{to}f\\ \begin{array}{c}\text{case-1}\text{(at position}2)\\ \begin{array}{cc}-\frac{y}{u}& =-8\\ v& =8u\end{array}\\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{f}=\frac{1}{-8u}+\frac{1}{-u}\\ f=\frac{-8u}{9}-\left(1\right)\end{array}\end{array}$

case -2 (at position 1 )

$\begin{array}{cc}\frac{-v}{u}& =-2\\ v_1& =2u_1\\ \frac{1}{f}& =\frac{1}{-2u_1}+\frac{1}{-u_1}\\ f& =\frac{-2u_1}{3}\\ f& =\frac{-2}{3}(u+5)-\left(2\right)\end{array}$

By equation $\left(1\right)$ and 2

$\begin{array}{cc}-\frac{84}{9}& =-\frac{2}{3}(u+5)\\ \frac{4u}{3}-u& =5\\ u& =15cm\\ f=-\frac{8u}{9}& =-\frac{8}{9}\left(15\right)\\ f& =-\frac{40}{3}cm\end{array}$

So D None of these is the required option