A liquid is placed in a hollow prism of angle 60o. If angle of the minimum deviation is 30o, what is the refractive index of the liquid?

  • Option 1)

    1.41

  • Option 2)

    1.50

  • Option 3)

    1.65

  • Option 4)

    1.95

 

Answers (1)

\delta _{min}= 2i-A@13954

Condition of maximum deviation -

i=e  

- wherein

\delta _{min}= 2i-A

\mu = \frac{\sin \left ( \frac{\delta _{m}+A}{2} \right )}{\sin \left ( A/2 \right )}

 

 angle of minimum deviation  \delta _{min}= 30

 

A = 60

\mu = \frac{\sin \left ( \frac{A+\delta _{m}}{2} \right )}{\sin \left ( A/2 \right )}  =     \frac{\sin \frac{60+30}{2} }{\sin 30}

\mu = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}  =  \sqrt{2}  or  \mu = 1.41


Option 1)

1.41

Correct

Option 2)

1.50

Incorrect

Option 3)

1.65

Incorrect

Option 4)

1.95

Incorrect

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