Answer:
r = 1.61 x 10^{11} m
Explanation:
energy radiated (H) = 2.7 x 10^31 W
surface temperature (T) = 11,000 k
assuming ε = 1 and taking σ = 5.67 x 10^{-8} W/m^{2}.K^{4}
we can find the radius of the star from the equation below
H = A x ε x σ x T^{4}
where area (A) = 4 x π x r^{2} (assuming it is a sphere)
therefore the equation becomes
H = 4 x π x r^{2} x ε x σ x T^{4}
2.7 x 10^31 = 4 x π x r^{2} x 1 x 5.67 x 10^{-8} x (11,000)^{4}
r =
r = 1.61 x 10^{11} m
Answer:
It can occur only when light is incident on an interface where the index of refraction on the other side is less.
Explanation:
When the light passes from a denser medium, with refractive index n1, to another less dense medium, with refractive index n2, the incident light beam is refracted in such a way that it is not able to cross the surface between both media, the light beam is fully reflected and completely confining in the optically denser medium through which it propagates. For this phenomenon to occur, it is necessary that the angle of the incident light beam with respect to the normal be greater than or equal to the critical incidence angle θc. The critical angle can be calculated as :