NUMERICAL 1 [2079 GIE ‘A’]

Calculate the polarizing angle for light travelling from water of refractive index 1.33 to glass of refractive index 1.53. [1] Ans: 49^o

Solution

Polarizing angle [latex](\theta_p)[/latex] =?

Refractive index of water [latex](\mu_w^a) = 1.33[/latex]

Refractive index of glass [latex](\mu_g^a) = 1.53[/latex]

So, refractive index of glass w.r.t. water [latex](\mu_g^w) = \mu_g^a \times \mu_a^w[/latex]

      = 1.53 x [latex]\frac{1}{1.33}[/latex]

      = 1.150

Now, we know that,

[latex]\mu_g^w = tan\theta_p[/latex]

 Or, 1.15 = tan[latex]\theta_p[/latex]

Or, [latex]\theta_p   = tan^{-1}(1.15)[/latex]

      = 48.991^o [latex]\approx 49^o[/latex].

NUMERICAL 2 [2079 GIE ‘A’]

The polarizing angle for a medium is 60^o. Calculate the velocity of light in the medium. Ans: [latex]1.73\times 10^8[/latex] m/s

Solution

Polarizing angle [latex](\theta_p)[/latex] = 60^o

Velocity of light in the medium (v) =?

We know that,

[latex]\mu = tan\theta_p[/latex]

= tan60^o

      = [latex]\sqrt{3}[/latex]

Now,

[latex]\mu = \frac{velocity\ of\ light\ in\ vacuum\ (c)}{velocity\ of\ light\ in\ medium\ (v)}[/latex]

Or, [latex]\sqrt{3} = \frac{3\times 10^8}{v}[/latex]

Or, [latex]v = \frac{3\times 10^8}{\sqrt{3}}[/latex]

= [latex]1.73\times 10^8[/latex] m/s.

NUMERICAL 3

A beam of light is incident at polarizing angle on a piece of transparent material of refractive index 1.62. What is the angle of refraction for the transmitted beam? Ans: 31.65^o

Solution

Given,

Refractive index [latex](\mu) = 1.62[/latex]

Angle of refraction (r) =?

We know that,

[latex]\mu = tan\theta_p[/latex]

Or, 1.62 = tan[latex]\theta_p[/latex]

Or, [latex]\theta_p = tan^{-1}(1.62)[/latex]

      = 58.314^o

Now,

[latex]\theta_p + r = 90^o[/latex]

Or, r = [latex]90^o – \theta_p[/latex]

      = 90 – 58.314

      = 31.686^o.

NUMERICAL 4

The critical angle for light in a certain substance is 45^o. What is the polarizing angle? Ans: 54.7^o

Solution

Critical angle (c) = 45^o

Polarizing angle [latex](\theta_p)[/latex] =?

We know that,

[latex]\mu = tan\theta_p[/latex]

Or, [latex]\frac{1}{sinc} = tan\theta_p\ [\because \mu = \frac{1}{sinc}][/latex]          

Or, [latex]\frac{1}{sin45^o} = tan\theta_p[/latex]

Or, [latex]\sqrt{2} = tan\theta_p[/latex]

Or, [latex]\theta_p = tan^{-1}\sqrt{2} = 54.74^o[/latex]

NUMERICAL 5

The polarizing angle for a transparent medium is 60^o. What is the refractive index of the medium? Ans: [latex]\sqrt{3}[/latex]

Solution

Polarizing angle [latex](\theta_p) = 60^o[/latex]

Refractive index [latex](\mu)[/latex] =?

We know that,

[latex]\mu = tan\theta_p[/latex]

            = tan60^o

                        = [latex]\sqrt{3}[/latex]

NUMERICAL 6

A parallel beam of unpolarized light in air is incident at an angle of 55^o on a plane glass surface. If the reflected beam is completely plane polarized, find the refractive index of the glass and the angle of refraction of the transmitted beam.

Solution

Given,

Angle of incidence [latex](i = \theta_p)[/latex] = 55^o

Refractive index of the glass [latex](\mu)[/latex] =?

Angle of refraction (r) =?

We know that,

[latex]\mu = tan\theta_p[/latex]

            = tan55^o

            = 1.428

Now, r + [latex]\theta_p = 90[/latex]

Or, r = 90 – [latex]\theta_p[/latex]

            = 90 – 55

            = 35^o.