Question

Refractive index of the solid $\left(\mu \right)$ in terms of ${\mu }_0$ and i is

• A. $\sqrt{{\mu }_0^2+si{n}^{2}i}$
• B. ${\mu }_0+si{n}^{2}i$
• C. $\sqrt{{\mu }_0^2-2si{n}^{2}i}$
• D. $\sqrt{{\mu }_0^2-si{n}^{2}i}$

Answer 1

Answer: (D)

$\sqrt{{\mu }_0^2-si{n}^{2}i}$

Using snell's law for refraction through the vertical surface of prism A,

$1\times sini={\mu }_o\times sinr$

Thus $cosr=\sqrt{1-(\frac{sini}{{\mu }_o}{)}^2}$ - (1)

Using snell's law for refraction through the horizontal surface ,

${\mu }_o\times sinC=\mu \times sin{90}^o$

Thus $sinC=\frac{\mu }{{\mu }_o}$ -(2)

Also $C+r={90}^o$

So, $sinC=cosr$

Using (1) & (2), $\frac{\mu }{{\mu }_o}=\sqrt{1-(\frac{sini}{{\mu }_o}{)}^2}$

Hence , $\mu =\sqrt{{\mu }_o^2-si{n}^{2}i}$