A thin convex lens L ( refractive index = 1.5  ) is placed on a plane mirror M. When a pin is placed at A , such that OA=18\:cm , its real inverted image is formed at A itself , as shown in figure . When a liquid of refractive index \mu _{l} is put between the lens and the mirror , the pin has to be moved to A' such that OA'=27\:\:cm , to get its inverted real image at A' itself . The value of \mu _{l} will be :

  • Option 1)

    \frac{4}{3}

  • Option 2)

    \frac{3}{2}

  • Option 3)

    \sqrt{3}

  • Option 4)

    \sqrt{2}

 

Answers (1)
S solutionqc

P_{initial} = 2P_{L} = \frac{1}{9} \Rightarrow P_{L} = \frac{1}{18}

P_{final}=2P_{L}+2P^{'} = \frac{2}{27}

2P^{'}=\frac{2}{27}-\frac{1}{9} \Rightarrow P^{'}=-\frac{1}{54}

-\frac{1}{54}=(\mu-1)\left ( \frac{1}{-18}-\frac{1}{\infty } \right )

\Rightarrow \mu=\frac{4}{3}

 


Option 1)

\frac{4}{3}

Option 2)

\frac{3}{2}

Option 3)

\sqrt{3}

Option 4)

\sqrt{2}

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