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Confused! kindly explain, - Optics - JEE Main-3

A ray of light AO in vacuum is incident on a glass slab at angle 60^{\circ} and

refracted at angle 30^{\circ} along OB as shown in the figure. The optical path length of light 

ray from A to B is : 

  • Option 1)

    \frac{2\sqrt3}{a}+2b

  • Option 2)

    2a+\frac{2b}{3}

  • Option 3)

    2a+\frac{2b}{\sqrt3}

  • Option 4)

    2a+2b

 
Answers (1)
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Optical Path -

x{}'=\mu \cdot x
 

- wherein

x{}'= Distance travelled in vacuum

x= Distance travelled in a medium of refractive index \mu

 

 

Optical path = AO + \mu ( OB )

Apply Snells law ,

1\cdot sin60^{\circ}=\mu \cdot sin30^{\circ}

\mu=\frac{\sqrt3}{2}/\frac{1}{2}=\sqrt3

sin 30^{\circ}=\frac{a}{AO}

So, AO = 2a

Similarly,

cos30^{\circ}=\frac{b}{OB}

=> OB=\frac{b}{cos30^{\circ}}

=> OB=\frac{2}{\sqrt3}b

So, optical path = AO + \mu ( OB )

                          = 2a+(\sqrt3\times \frac{2}{\sqrt3}b)

                         = 2a+2b

    

 


Option 1)

\frac{2\sqrt3}{a}+2b

Option 2)

2a+\frac{2b}{3}

Option 3)

2a+\frac{2b}{\sqrt3}

Option 4)

2a+2b

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