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.
Electrical charge are measured in columbs, usually demoted as C. Hence, the charges on proton and electron will be measured in Coloumbs. It typically measures the amount of electricity conveyed per second by a current of 1 ampere. The other units Given such as ; Volt is used for measuring voltage, which is the pressure in an electrical source. AMPERE is used for measuring the current flowing through an electrical circuit.
Dalton is a unit of mass and is about 1.660 * 10^-27 kg
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.