Question

The velocity of light in water is 1.5×10 to the power of 8 m/s what is the polarizing angel of incidence

Answer 1

Answer:

we know that refractive index n=

speedoflightinmedium

speedoflightinvacuum

where c=3×10

8

m/s

1.5=

v

c

v=

1.5

3×10

8

v=2×10

8

m/s

For critical angle we will use snell's law

C is critical angle

n

glass

sinC=n

air

sin90 n

air

=1

sinC=

n

glass

1

=

1.5

1

C=41.8

o

Answer 2

The polarizing angle of incidence is 30°.

Given:

Velocity of light in water $=1.5$ × $10^{8}m/s$

To find:

Polarizing angle of incidence.

Solution:
We know,

When a ray of light traveling through a medium passes towards another medium, the ray of light undergoes refraction in a way such that the ray bends towards or way from the normal at the point of incidence.

The refractive index μ of the medium here, is given by the ratio of speed of light in vacuum $(c)$ to the speed of light in the medium $(v)$.

                               μ $=\frac{c}{v}$

At a certain angle of incidence the ray will be projected in such a way that there will be reflected ray as well as refracted ray both particularly perpendicular to each other. This incident angle is called as the polarizing angle or Brewster's angle $(i_{b} )$.

                            μ   $=\frac{1}{sin(i_{b}) }$

From the two equations of refractive index, we get

   μ $=\frac{c}{v} =\frac{1}{sin(i_{b} )}$

We know,

$c=3$ × $10^{8}m/s$

$v=1.5$ × $10^{8}m/s$

Substituting the given values in the equation, we get

$\frac{3*10^{8} }{1.5*10^{8} }= \frac{1}{sin(i_{b} )}$

$sin(i_{b} )=\frac{1}{2}$

$i_{b} =30^{o}$

Final answer:

Hence, the polarizing angle of incidence is 30°.