Question

if an object is placed 6 cm from a concave mirror the image distance is 12cm what is the image distance if the object is moved from the original position 1 cm away from the mirror ?

Answer 1

Dear Student,
Using the mirror formula for concave mirror,
$$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v} \\ u=-6cm \\ v=-12cm \\ Thus, \\ \frac{{\displaystyle 1}}{{\displaystyle f}}=-\frac{{\displaystyle 1}}{{\displaystyle 6}}-\frac{{\displaystyle 1}}{{\displaystyle 12}}=\frac{-3}{12}=-\frac{{\displaystyle 1}}{{\displaystyle 4}} \\ So,focallengthofthecocavemirror,f=-4cm \\ Nowtheobjectismoved1cmawayfromthemirror,so \\ newobjectdis\tan ce,u\text{'}=-(6+1)=-7cm \\ newimagedis\tan ce=v\text{'} \\ U\sin gthemirrorformula \\ \frac{{\displaystyle 1}}{{\displaystyle f}}=\frac{{\displaystyle 1}}{{\displaystyle u\text{'}}}+\frac{{\displaystyle 1}}{{\displaystyle v\text{'}}} \\ \frac{{\displaystyle 1}}{{\displaystyle v\text{'}}}=\frac{{\displaystyle 1}}{{\displaystyle f}}-\frac{{\displaystyle 1}}{{\displaystyle u\text{'}}}=-\frac{1}{4}+\frac{1}{7}=\frac{-7+2}{14}=\frac{{\displaystyle -5}}{14} \\ v\text{'}=-\frac{14}{5}=-2.8cm \\ Imagewillbeformed2.8cmawayfromthemirroristhesamesideoftheobject. \end{gathered}$$
Regards