Question

a object is placed at the distance of 12 cm from a radius ofcurvature 16 cm find the position of image
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Answer 1

Answer:

$\huge\mathsf\red{Given}$

$\mathsf{Distance(u) \: = \: -12 \: cm}$

$\mathsf{Radius \: of \: curvature \: = \: -16 cm}$

$\huge\mathsf{f \: = \: \frac{R}{2} \: = \: \frac{-16}{2} \: = \: -8}$

$\huge\mathsf{\frac{1}{v} \: = \: \frac{1}{f} \: - \: \frac{1}{u}}$

$\huge\mathsf{\frac{1}{v} \: = \: \frac{1}{-8} \: - \: \frac{1}{-12}}$

$\huge\mathsf{\frac{1}{v} \: = \: \frac{1}{-8} \: + \: \frac{1}{12}}$

$\huge\mathsf{\frac{1}{v} \: = \: \frac{-3+2}{24}}$

$\huge\mathsf{\frac{1}{v} \: = \: \frac{-1}{24}}$

$\fbox\blue{v \: = \: 24 \: cm}$

$\mathsf\red{the \: image \: is \: virtual, \: erect, \: magnified \: and}$ $\mathsf\red{ is \: formed \: at \: the \: back \: of \: the \: mirror}$