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Messages: 1 - 3 of 3
• #### Message 1.

Posted by U15737416 (U15737416) on Sunday, 26th May 2013

Hi

I've been going through past papers and I know its a little late but I need help understanding one question. Its Higher physics 2011 Q27aiiB and I really don't understand it. I've looked at the answer and I can't understand how to get there. If anyone would be willing to help it would be greatly appreciated

Thanks

• #### Message 2

, in reply to message 1.

Posted by BitesizePhysicsTeacher (U15165682) on Sunday, 26th May 2013

Because the ray is emerging along the boundary, it must meet the boundary at the critical angle θc.

sin θc = 1/n --> θc = 37°

Angle X is the angle between the incident ray and the internally reflected ray, so as angle i = angle r, it must be TWICE the angle of incidence = 74°

• #### Message 3

, in reply to message 2.

Posted by U15737416 (U15737416) on Sunday, 26th May 2013

Because the ray is emerging along the boundary, it must meet the boundary at the critical angle θc.

sin θc = 1/n --> θc = 37°

Angle X is the angle between the incident ray and the internally reflected ray, so as angle i = angle r, it must be TWICE the angle of incidence = 74°
That makes so much sense thank you

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