Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let denote the speed of light in the air. Assume that the light entered the boundary at an angle of to the normal and exited with an angle of . By Snell's Law, the sine of and would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now . (in this case, .)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For , is monotonically increasing with respect to . In other words, for in that range, the value of increases as the value of increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle would have a larger . such that light would travel faster in that medium.
Ohm's law states that the electrical current (I) flowing in an circuit is proportional to the voltage (V) and inversely proportional to the resistance (R). Therefore, if the voltage is increased, the current will increase provided the resistance of the circuit does not change.
In the metals the electrons are the charge carriers and they are the only mobile charges causing the flow of current when subjected to potential difference.
The electrons being negatively lead to the flow of current in opposite direction. The direction of the current is opposite to the direction of the negative charges.