Here,
Size of object = 1.5 m;
Object distance (u) = - 2m = -200cm
focal length (f) = - 3.0 cm
Image distance (v) = ?
Using lens formula
[tex]\begin{gathered} \frac{1}{f}=\text{ }\frac{1}{v}-\frac{1}{u}; \\ \frac{1}{-3}=\frac{1}{v}-\frac{1}{-200}; \\ \frac{1}{-3}=\text{ }\frac{1}{v}\text{ +}\frac{1}{200} \\ \frac{1}{v}=\text{ -}\frac{1}{3}-\frac{1}{200}=-\text{ }\frac{203}{600} \\ v=\text{ - }\frac{600}{203}=\text{ -2.96 cm} \end{gathered}[/tex]Now magnification is given by
[tex]Magnification=\text{ }\frac{v}{u}=\text{ }\frac{2.96}{200}=0.0148[/tex]Final answer is :-
Image distance = - 2.96 cm & magnification = 0.0148
