A ray in glass is incident onto a water-glass interface, at an angle of incidenc

Question

A ray in glass is incident onto a water-glass interface, at an angle of incidence equal to half the critical angle for that interface. The indices of refraction for water and the glass are 1.33 and 1.43, respectively. The angle that the refracted ray in the water makes with the normal is closest to

The correct answer is 37 degrees. Please explain why.

QUESTION 2

A tank holds a layer of oil, 1.43 m thick, which floats on a layer of syrup that is 0.64 m thick. Both liquids are clear and do not intermix. A ray, which originates at the bottom of the tank on a vertical axis, crosses the oil-syrup interface at a point 0.90 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m from the axis and at the critical angle. The index of refraction of the oil is closest to

The answer is 1.64. Please explain!

Explanation / Answer

1)

iven,

i = 1/2(critical angle),

?1= 1.33, ?2= 1.43,

i C (critical angle)

                         =sin-(?1/?2)

                         = sin-1(1.33/1.43)

                           = 68.446o

incidence angle = 1/2(critical angle)

                         = 1/2(68.446) =34.223o

refracted angle = sin-(?2sin(i)/?1)

                            sin-(1.43*sin(34.223)/1.33)

                             =37.2074o (refracted ray in the water makes with the normal),

2)

This is a really simple question. A lot of unwanted data are given.

Take the tangent of the angle of incidence (a) at the oil-air interface.

We know the opposite side length for that angle is (2-0.9) = 1.1m

The adjacent side = 1.43m

So, tan(a)=1.1/1.43 = 0.77

Hence, a = 37.56 degrees

So, sin(a) = 0.61

Since a is the critical angle for oil-air interface, sin(a) will be the ratio of refractive indices of the materials

sin(a) = n(air) / n(oil)

Since n(air) = 1,

n(oil) = 1/sin(a) = 1/0.61 = 1.64..

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