Answer:
» Image distance :
[tex]{ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\ [/tex]
[tex]{ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}[/tex]
» Magnification :
• Let's derive this formula from the lens formula:
[tex] { \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\ [/tex]
» Multiply throughout by fv
[tex]{ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}[/tex]
• But we know that, v/u is M
[tex]{ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}[/tex]
[tex]{ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}[/tex]
» Nature of Image :
