Question

help me to solve this question i​

Answer 1

Answer:

$\sf \star$ From mirror formula:

$\sf \implies \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \\ \sf \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ \\ \sf \implies \frac{1}{v} = \frac{ u - f}{fu} \\ \\ \sf \implies v = \frac{fu}{u - f} \ \ \ \ \ \ \ \ ...eq_{1}$

$\sf \star$ Magnification of special mirror:

$\sf \implies m = - \frac{v}{u} \\ \\ \sf Substituting \ value \ of \ v \ from \ eq_{1}, we \ get: \\ \sf \implies m = - ( \frac{ \frac{fu}{u - f} }{u} ) \\ \\ \sf \implies m = - ( \frac{f \cancel{u}}{ \cancel{u}(u - f)} ) \\ \\ \sf \implies m = - \frac{f}{u - f} \\ \\ \sf \implies \boxed{ \bold{m = \frac{f}{f - u} }}$

Symbols in the answer have there usual meaning i.e.

u $\sf \rightarrow$ Image distance

v $\sf \rightarrow$ Object distance

f $\sf \rightarrow$ Focal length

m $\sf \rightarrow$ magnification

Answer 2

Answer:

1/v + 1/u = 1/f

1/v = 1/f - 1/u

v = fu/u - f

We know:

m = -v/u

m = -fu/u(u - f)

m = -f/(u - f)

m = f/f - u