Question

This question has multiple correct options

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Updated on : 2022-09-05

Solution

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Correct options are A) and B)

90 $−θ<$ sin $_{−1}(43 )$

$θ>2π −$ sin $_{−1}43 $

Similarly if $θ>90_{∘}$ , then

for No TIR

$θ<2π +$ sin $_{−1}43 $

so $(2π −sin_{−1}43 <θ<2π +sin_{−1}43 )$

For calculating focal length of lens in air

$f1 =(μ_{rel}−1)(R_{1}1 −R_{2}1 )$

$f1 =(μ_{rel}−1)(R_{1}1 −R_{2}1 )$

$501 =(23 −1)(k)⇒k=251 $

$f_{w}=4f_{a}=4×50cm=200m$

$f_{w}1 =(23 43 −1)(k)=8k $

$⇒f_{w}=200cm$

So that final image will be formed 200 cm in right of lens.

$⇒Imagedistance,OI=(100+200)cm=300cm$

Ans. (a) and (b)$(2π −sin_{−1}43 <θ<2π +sin_{−1}43 )$

(d) OI = (100 + 200) cm = 300 cm but this distance is valid only when $θ$ is between $θ_{min}$ to $θ_{max}$

Ans. (a) and (b) $(2π −sin_{−1}43 <θ<2π +sin_{−1}43 )$

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