Question

Glass has refractive index u with respect to air and the critical angle for a ray of light going from glass to air is $\theta$. If a ray of light is incident from air on the glasses with the angle of incidence $\theta$, the corresponding angle of refraction is

• A. ${\sin }^{-1}\left(\frac{1}{\sqrt{u}}\right)$
• B. ${90}^0$
• C. ${\sin }^{-1}\left(\frac{1}{{\mu }^2}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{\mu }\right)$

Answer 1

Answer: (C)

${\sin }^{-1}\left(\frac{1}{{\mu }^2}\right)$

As we know,,

${}_a{\mu }_g=\frac{1}{sin\theta }\Rightarrow \mu =\frac{1}{sin\theta }...\left(i\right)$

Now from Snell's law $\mu =\frac{sini}{sinr}=\frac{sin\theta }{sinr}$

$\Rightarrow sinr=\frac{sin\theta }{\mu }....\left(ii\right)$

From equation (i) and (ii)

$sinr=\frac{1}{{\mu }^2}\Rightarrow r=si{n}^{-1}\left(\frac{1}{{\mu }^2}\right)$