Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC.  At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 900.  Angle A is given by :

 

  • Option 1)

    \frac{1}{\cos^{-1} \left ( \sin \Theta c \right )}

  • Option 2)

    \frac{1}{\tan ^{-1}\left ( \sin \Theta c \right )}

  • Option 3)

    \cos ^{-1}\left ( \sin \Theta c \right )

  • Option 4)

    \tan ^{-1}\left ( \sin \Theta c \right )

 

Answers (1)

As we learnt in

Relation between angle of incidence and angle of refaction -

\mu _{1}\sin i= \mu _{2}\sin r
 

- wherein

\mu _{1}= refractive index of medium of incidence.

\mu _{2}=  refractive index of medium where rays is refracted.

i= angle of incidence.

r= angle of refraction.

 

 

 

 From Snell's law 

\frac{\mu_{R}}{\mu_{D}}=\frac{sini}{sinr}                                (i)

Where \angle i=A and \angle r=(90-A)

We know that sin\theta_{c}=\frac{\mu_{R}}{\mu_{D}}

\Rightarrow\ \; sin\theta_{c}=\frac{sinA}{sin(90-A)}=\frac{sinA}{cosA}

    sin\theta_{c}=tanA

\Rightarrows\ \;tanA^{-1}(sin\theta_{c})

Correct option is 4.

 

 


Option 1)

\frac{1}{\cos^{-1} \left ( \sin \Theta c \right )}

This is an incorrect option.

Option 2)

\frac{1}{\tan ^{-1}\left ( \sin \Theta c \right )}

This is an incorrect option.

Option 3)

\cos ^{-1}\left ( \sin \Theta c \right )

This is an incorrect option.

Option 4)

\tan ^{-1}\left ( \sin \Theta c \right )

This is the correct option.

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