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Light enters a substance from air at 45.0 degrees to the normal.  It continues through...

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lkehoe | Valedictorian

Posted May 19, 2013 at 5:47 AM via web

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Light enters a substance from air at 45.0 degrees to the normal.  It continues through the substance at 34.7 degrees to the normal.  What would be the critical angle for this substance?

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pramodpandey | College Teacher | Valedictorian

Posted May 19, 2013 at 6:20 AM (Answer #1)

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We know that

`n_1sin(theta_1)=n_2sin(theta_2)`

where `theta_1`  is the angle subtended between the incident ray and the normal to the interface, and `theta_2`  is the angle subtended between the refracted ray and the normal to the interface. The quantities `n_1`  and `n_2`  are termed the refractive indices of media 1(air) and 2 (substance), respectively.

`theta_1=45^o , n_1=1.0`

`theta_2=34.7^o ,n_2=?`

`n_2=(sin(45^o))/(sin(34.7^o))`

`=1.2421`

Also

`sin(I_c)=1.0/n_2`

`I_c=sin^(-1)(1.0/n_2)`

`=sin^(-1)(1/1.2421)`

`=53.62^o`

The critical angle for this substance=53.62 degree.

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