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Sagot :
According to problem ( considering sign convention for convex lens)
height of object = 2.25 mm;
object distance(u)= - 8.5 cm
focal length (f)= 5.5 cm
image distance(v)= ?
Using lens formula
[tex]\begin{gathered} \frac{1}{f}=\text{ }\frac{1}{v}-\frac{1}{u}; \\ \therefore\frac{1}{5.5}=\text{ }\frac{1}{v}\text{ -}\frac{1}{-8.5} \\ \frac{1}{5.5}=\frac{1}{v}\text{ +}\frac{1}{8.5}; \\ \frac{1}{v}=\frac{1}{5.5}-\frac{1}{8.5}=\frac{3}{5.5\times8.5}=\frac{3}{46.75} \\ v=\frac{46.75}{3}=15.58\text{ cm} \end{gathered}[/tex]a) Answer is :- Location of image = 15.58cm
Using magnification formula
[tex]\begin{gathered} Magnification\text{ =}\frac{height\text{ }of\text{ }image}{height\text{ of object}}=\frac{v}{u} \\ \therefore Magnification=\text{ }\frac{15.58}{8.5}\text{ =1.83} \\ \therefore\frac{height\text{ }ofimage}{2.25}\text{ =1.83;} \\ \therefore height\text{ of image = 1.83}\times2.25=4.12\text{ mm} \end{gathered}[/tex]b) Height of image = 4.12 mm
C) Magnification= 1.83
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