The speed of light in media ' A ' and ' B ' are 2 . 0 × 10 ^ 10 cm...

Given:${v}_{A} = 2 \times {10}^{8}m{s}^{- 1}$ and ${v}_{B} = 1 . 5 \times {10}^{8}m{s}^{- 1}$

For critical angle:$\sin \left({\theta}_{c}\right) = \dfrac{{\mu}_{A}}{{\mu}_{B}} = \dfrac{c}{{v}_{A}} \times \dfrac{{v}_{B}}{c} = \dfrac{1 . 5 \times {10}^{8}}{2 \times {10}^{8}} = \dfrac{3}{4}$

$⇒ {\theta}_{c} = \sin^{- 1}\left(\dfrac{3}{4}\right)$

Now for $TIR, \theta > {\theta}_{C}$

So, ${\theta}_{c} > \sin^{- 1}\left(\dfrac{3}{4}\right)$