When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted light, and all the light is reflected. The value of this angle is given by

where n2 and n1 are the refractive indices of the second and first medium, respectively.
In the first part of the problem, light moves from glass to air (

) and the critical angle is

. This means that we can find the refractive index of glass by re-arranging the previous formula:

Now the glass is put into water, whose refractive index is

. If light moves from glass to water, the new critical angle will be

In order to find it's square root, we could make it into two square roots.

Let us find the square roots of both radicals seprately.

Each pair of a number inside square root gives a number out .



Therefore,


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There is no adjustment in gravity, yet there is an adjustment in 'weightness'.
Gravitational compel and weight with respect to an edge are not similar things, despite the fact that it is normally educated something else.
Weight is really the aggregate of gravitational powers and of inertial drive for a question very still (no Coriolis compel) in a given casing.
In the event that the Earth were not pivoting, weight would increment most at the Equator and be unaltered at the Poles.