Question #7abf9

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1s2s2p
Dec 28, 2017

#"Critical angle"=pi/6=30^circ#

Explanation:

The equation for critical angle is:
#sinC=n_1/n_2#, where:

  • #C# = critical angle
  • #n_1# = boundary medium
  • #n_2# = initial medium

In this case #n_1# is the vacuum as it is the medium acting as a boundary, while #n_2# is the other medium, as the light is travelling through that medium.

#n_1=1#

#sinC=1/2#

#C=sin^(-1)(1/2)=pi/6=30^circ#

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