An incident beam of light moves through liquid water ( 1.33) and impinges upon an interface with glass ( 1.52) with an angle of incidence of 47.0o. What is the angle of the refracted beam (i. e., in the glass) with respect to the surface normal

The angle of the refracted beam with respect to the surface normal is .

Explanation:

The direction of a beam of light changes when it crosses the interface between two media with different index of refraction. We will consider that n₁ and n₂ are the index of refraction of the first and the second medium and θ₁  and θ₂ are the angles that the beam of light makes with the surface normal in the first and the second medium respectively.

Snell's Law relates the angle of incidence θ₁ and the angle of refraction θ₂ of a beam of light with the index of refraction in each medium:

We are told that:

if we rearrange the equation:

we get an expression for the angle of the refracted beam:

replacing the values we get that:

finally we get that:

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft

did yall get answers for this yet