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  • Help me please, A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from leftto right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from

A planoconvex lens becomes an optical system of 28 cm focal length when its plane surface is silvered and illuminated from left
to right as shown in Fig-A. If the same lens is instead silvered on the curved surface and illuminated from other side as in Fig.B, it acts like an optical system of focal length 10 cm. The refractive index of the material of lens is :

  • Option 1)

    1.50

  • Option 2)

    1.55

  • Option 3)

    1.75

  • Option 4)

    1.51

 

Answers (2)

best_answer

As we learned 

 

Lens placed close to each other -

\frac{1}{f_{eq}}= \frac{1}{f_{1}}+ \frac{1}{f_{2}}------+ \frac{1}{f_{n}}

\frac{1}{f_{eq}}=Equivalent focal length.

 

- wherein

f_{1},f_{2},------f_{n} are focal lenght of lenc 1, 2, 3, -----n

 

  

p=2p_{1}+p_{2} = 2\left ( \frac{\mu -1}{R} \right )+\frac{2}{R}=\frac{1}{28}+\frac{2}{R}

\Rightarrow \frac{2}{R}+\frac{1}{28}=\frac{1}{10} \Rightarrow \frac{2}{R}=\frac{1}{10}-\frac{1}{28} = \frac{18}{280}

R = \frac{280}{9}cm

\because \frac{2(\mu -1)}{R} = \frac{1}{28} \Rightarrow \mu -1= \frac{280}{9\times 2}\times \frac{1}{28}

\Rightarrow \mu =\frac{14}{9} = 1.55


Option 1)

1.50

Option 2)

1.55

Option 3)

1.75

Option 4)

1.51

Posted by

Avinash

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