Question

A ray of light while travelling from a denser to a rarer medium undergoes total reflection. Derive the expression for the critical angle in terms of the speed of light in the respective media.

Answer 1

Answer:

$\sin(i _{c}) = \dfrac{v _{2}}{v _{1}}$

Explanation:

Ler a ray trying to enter medium 2 (rarer) from medium 1 where (denser) such that total internal refection occurs,

The angle of incidence is same as critial angle and

angle of refraction = 90°

(consider the attachment).

Using snell's law,

$\dfrac{n_{1}}{n_{2}} = \dfrac{ \sin(i_{c}) }{ \sin(r) } \\ \dfrac{ \sin(i_{c}) }{ \sin(90) } = \dfrac{n_{1}}{n_{2}} \\ \sin(i_{c}) = \dfrac{n_{1}}{n_{2}}$

Using

$n = \dfrac{c}{v}$

Where c = speed of light in vaccum and v = speed of light in a medium with refractive index n.

$\sin(i _{c}) = \dfrac{( \dfrac{c}{v _{1}} )}{( \dfrac{c}{v _{2} }) } = \dfrac{v _{2}}{v _{1}}$

Which is our final expression.

Answer 2

Answer:

The critical angle in terms of the speed of light in the respective media is $i_c=(sin)^-^1\frac{v_2}{v_1}$.

Explanation:

When a ray of light travels from medium2 (dense) to medium1 (rarer), total internal reflection occurs.

Now by using  snell's law,

$\mu_1 sin r = \mu_2 sin i_c$        (1)

Where,

μ₁=refractive index of the rarer medium

μ₂=refractive index of the denser medium

$i_c$=critical angle of incidence

r=angle of refraction

Now since total internal reflection is occurring here so the angle of refraction(r) = 90°

We get the following results by changing the value of "r" in equation (1):

$\mu_1 sin 90\textdegree = \mu_2 sin i_c$

$sini_c=\frac{\mu_1}{\mu_2}$      (2)

Also, we know that,

$\mu=\frac{c}{v}$      (3)

μ=refractive index of a medium

c=speed of light in a vacuum

v=speed of light in a particular medium

As a result, equation (2) can be written as,

$sini_c=\frac{\frac{c}{v_1} }{\frac{c}{v_2}}$

$sini_c=\frac{v_2}{v_1}$

$i_c=(sin)^-^1\frac{v_2}{v_1}$

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