Question

Prove ${n_{23}} = {n_{13}} \times {n_{21}}$ from ${n_{12}} = \dfrac{1}{{{n_{21}}}}$

Answer 1

Snell’s Law of Refraction has to be used here. This law indicates the ratio of the sine of angles between the incidence and refraction of a wave is constant when it passes between two given media which can then be equated to medium -2 and medium -3 for different indexes and velocities. We can deduce the above equation from this method for the given index and velocity.

Complete step by step answer:
By Using the Snell’s Law of Refraction Formula we get
${n_1}\sin {\theta _1} = {n_2}\sin {\theta _2}$
Where ${n_1}$ = incident index,
${n_2}$ =refractive index
${\theta _1}$,${\theta _2}$ are the angles of incidence and the refracted angle respectively.
Hence $\dfrac{{{n_1}}}{{{n_2}}} = \dfrac{{\sin {\theta _2}}}{{\sin {\theta _1}}} = \dfrac{{{v_1}}}{{{v_2}}}$
Where ${v_1}$ = velocity of light in medium -1,
${v_2}$ = velocity of light in medium -2
Here we have ${n_{12}} = \dfrac{1}{{{n_{21}}}}$
where ${n_{21}} = \dfrac{{\sin {n_2}}}{{\sin {n_1}}}$
Similarly we can solve by applying the Snell’s law for medium -3
${n_{23}} = \dfrac{{{n_3}}}{{{n_2}}}$...... equation (1)
Whereas ${n_{13}} = \dfrac{{{n_3}}}{{{n_1}}}$...... equation (2)
${n_{12}} = \dfrac{1}{{{n_{21}}}}$ which implies ${n_{21}} = \dfrac{1}{{{n_{12}}}} = \dfrac{{{n_1}}}{{{n_2}}}$......equation (3)
Hence from equation (1),(2) and( 3) we get
$\dfrac{{{n_3}}}{{{n_2}}} = \dfrac{{{n_3}}}{{{n_1}}} \times \dfrac{{{n_1}}}{{{n_2}}}$
Hence LHS = RHS ${n_{23}} = {n_{13}} \times {n_{21}}$.

Note: Using Snell's Law of Refraction which indicates the ratio of angles between the sine of angles of incidence and refraction of a wave is constant when it passes between two given media. We can get the above theorem proved. Hence Snell's law of refraction predicts that the light ray always deviates more towards normal in the optically denser medium (medium of higher refractive index) and away from the normal in the optically rarer medium (medium of lower refractive index).