Answer from photo no. 10

1.	Answer from photo no. 10
Answer» $\sf{\large{\underline{\underline{EXPLANATION:}}}}$ $\textbf{\underline{Given,,}}$ Height of object,h=2.5mm=+0.25cm Height of image,h'=10mm=-1cm Distance of the object from the MIRROR,u=-5cm Distance of image from the mirror,v=?? FOCAL length,f=?? $\textbf{\underline{Using,,}}$ $\sf{\boxed{\frac{h'}{h}=\frac{-v}{u}}}$ $\textbf{\underline{We get,,}}$ $\sf{\to{\large{\frac{1}{0.25}=\frac{-v}{-5}}}}$ $\sf{\to{\large{4=\frac{v}{5}}}}$ $\sf{\to{\large{v=20cm}}}$ $\textbf{\underline{Using,,}}$ $\sf{\boxed{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}}}$ $\textbf{\underline{Substituting value of 'v' and 'u' in formula, we get,,}}$ $\sf{\to{\large{\frac{1}{20}+\frac{1}{-5}=\frac{1}{f}}}}$ $\sf{\to{\large{\frac{1}{20}-\frac{1}{5}=\frac{1}{f}}}}$ $\sf{\to{\large{\frac{1-4}{20}=\frac{1}{f}}}}$ $\sf{\to{\large{\frac{-3}{20}=\frac{1}{f}}}}$ $\sf{\to{\large{\frac{-20}{3}=f}}}$ $\sf{\to{\underline{\boxed{f=-6.67 cm}}}}$ HENCE,,Focal length,f=-6.67 cm $\rule{200}2$ HOPE it helps ❣️❣️

Answer»

$\sf{\large{\underline{\underline{EXPLANATION:}}}}$