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

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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?

Posted by lkehoe on May 19, 2013 at 5:47 AM via web and tagged with critical angle, degrees to the normal, light, optics, physics, science, substance

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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.

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

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