Here,
object distance(u)= - 3.00cm;
focal length (f)= -8.00cm
object distance(v)= ?
Using lens formula
[tex]\begin{gathered} \frac{1}{f}\text{ = }\frac{1}{u}\text{ + }\frac{1}{v};\text{ } \\ \frac{1}{8.00}\text{ = }\frac{1}{3.00}\text{ +}\frac{1}{v}; \\ \therefore\frac{1}{v}=\text{ }\frac{1}{8}\text{ -}\frac{1}{3}=\text{ - }\frac{5}{24} \\ \therefore v=-\frac{24}{5}\text{ = -4.8 cm} \end{gathered}[/tex]Note: Here object is placed between focus and pole of the mirror as
u< f. Therefore a real and inverted image will be formed in same side that of object . So image distance have negative sigh according to sign convention.
Final answer is - 4.8 cm
