Physics - Geometrical Optics Question with Solution | TestHub

From Snell's law

$\mathrm{\mu }=\frac{1}{\sin i_C}$

Since,$i_B >i_A$

$\sin i_{:B}>\sin i_A$

$\Rightarrow {\mathrm{\mu }}_B<{\mathrm{\mu }}_A$

When angle of incidence in the denser medium is increased even very slightly beyond the critical angle, then the ray of light is reflected back completely in the denser medium and total internal reflection takes place. Let$\mathrm{\theta }$be critical angle from medium$A \mathrm{t}\mathrm{o}\mathrm{ }B,$then

${ }_B{\mathrm{\mu }}_A=\frac{1}{\sin \mathrm{\theta }}$

$\Rightarrow \sin \mathrm{\theta }=\frac{1}{{ }_B{\mathrm{\mu }}_A}\frac{{\mathrm{\mu }}_B}{{\mathrm{\mu }}_A}$

$=\frac{1}{\sin i_B}\times \frac{\sin i_A}{1}$

$\Rightarrow \mathrm{\theta }={\sin }^{-1}\left[\frac{\sin i_A}{\sin i_B}\right]$