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Sagot :
Here ,
object distance (u)= -1.00 cm
image distance(v)= -10.0 cm
focal length = -f
radius of curvature =r
Using concave mirror formula
[tex]\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{v}; \\ \therefore\frac{1}{-f}=\frac{1}{-1}+\frac{1}{-10}; \\ \frac{1}{f}=\text{ }\frac{11}{10}; \\ \therefore f=\text{ }\frac{10}{11}; \end{gathered}[/tex]Again we have
r= 2f
[tex]\begin{gathered} r=\text{ 2f } \\ \therefore r=\text{ 2}\times\frac{10}{11}\text{ =1.818 cm = 1.82cm\lparen Approx\rparen} \end{gathered}[/tex]answer is 1.82 cm ( negative in sign)
Now magnification is
[tex]magnification\text{ = }\frac{v}{u}=\text{ }\frac{10}{1}\text{ = 10 }[/tex]
Answer is radius = 1.82cm( negative in sign) & magnification =10
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