Consider an optical system placed in water $\left(\mu =4/3\right)$ PQRS is a hollow glass slab filled with air $\left(\mu =1\right)$ This slab is kept on the principle axis of a lens in such a way that surface PS makes an angle $\theta$ with the axis of the lens. $\left(0<\theta <180^{\circ }\right)$, then

$if\theta <90^{\circ }$ and If there is no TIR. 

90 $-\theta <$ sin ${}^{-1}\left(\frac{3}{4}\right)$

$\theta >\frac{\pi }{2}-$ sin ${}^{-1}\frac{3}{4}$

Similarly if $\theta >90^{\circ }$ , then

for No TIR

$\theta <\frac{\pi }{2}+$ sin ${}^{-1}\frac{3}{4}$

so $\left(\frac{\pi }{2}-si{n}^{-1}\frac{3}{4}<\theta <\frac{\pi }{2}+si{n}^{-1}\frac{3}{4}\right)$

For calculating focal length of lens in air
$\frac{1}{f}=\left({\mu }_{rel}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{50}=\left(\frac{3}{2}-1\right)\left(k\right)\Rightarrow k=\frac{1}{25}$

$f_w^g=4f_a^G=4\times 50cm=200m$

$\frac{1}{f_w^o}=\left(\frac{3}{2}\frac{3}{4}-1\right)\left(k\right)=\frac{k}{8}$

$\Rightarrow f_w=200cm$

So that final image will be formed 200 cm in right of lens.

$\Rightarrow Imagedistance,OI=(100+200)cm=300cm$

Ans. (a) and (b)$\left(\frac{\pi }{2}-si{n}^{-1}\frac{3}{4}<\theta <\frac{\pi }{2}+si{n}^{-1}\frac{3}{4}\right)$

(d) OI = (100 + 200) cm = 300 cm but this distance is valid only when $\theta$ is between ${\theta }_{min}$ to ${\theta }_{max}$ 

Ans. (a) and (b) $\left(\frac{\pi }{2}-si{n}^{-1}\frac{3}{4}<\theta <\frac{\pi }{2}+si{n}^{-1}\frac{3}{4}\right)$