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Question

A biconves of glass of refractive index 1.51.5. Find its focal length. What should be the value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass?
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Solution

$$\dfrac { 1 }{ f_m}=(\dfrac { \mu_g }{ \mu_m } -1)$$$$(\frac { 1 }{ R_1 }-\dfrac { 1 }{R_2  } )$$ .....1
When lens in air $$\dfrac { 1 }{ f_a}=( \mu_g -1)$$$$(\dfrac { 1 }{ R_1 }-\dfrac { 1 }{R_2  } )$$ .....2
From equation (i) and (ii)
$$\dfrac {  f_a}{ f_m } =\dfrac { \dfrac {\mu_g  }{\mu_m  }-1 }{(\mu_g-1)  } $$
$$\dfrac {  20cm}{ f_m } =\dfrac { \dfrac {1.5}{1.65  }-1 }{(1.5-1)  } $$

Physics

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