Physics - Geometrical Optics Question with Solution | TestHub

P: Let $i_1$ be the angle of incidence on AC, and $r_1$ be the angle of refraction. The angle of incidence on face AB is $r_2=45^{\circ }-r_1$. For TIR at AB, $r_2>C$, where

$C={\sin }^{-1}(1/\mu )={\sin }^{-1}(1/1.5)={\sin }^{-1}(2/3)\approx 41.81^{\circ }$. So, $45^{\circ }-r_1>C\Rightarrow r_1<45^{\circ }-C\approx 3.19^{\circ }$. Using Snell's law at AC:

$1\cdot \sin i_1=\mu \cdot \sin r_1=1.5\sin (45^{\circ }-{\sin }^{-1}(2/3))$.

This evaluates to $\sin i_1=(\sqrt{10}-2\sqrt{2})/4\approx 0.0835$, so $i_1\approx 4.8^{\circ }$.

Q: When immersed in water, the critical angle $C^{\prime }={\sin }^{-1}({\mu }_W/{\mu }_{prism})={\sin }^{-1}((4/3)/1.5)={\sin }^{-1}((4/3)/(3/2))={\sin }^{-1}(8/9)$.

R: If light is incident normally on face AB, it enters undeviated. The angle of incidence on face AC is $45^{\circ }$. The critical angle for $\mu =\sqrt{2}$ is $C={\sin }^{-1}(1/\sqrt{2})=45^{\circ }$. Since the angle of incidence ($45^{\circ }$) equals the critical angle, the light grazes the surface AC. The ray emerges along AC, deviating by $45^{\circ }$ from its original direction (which was perpendicular to AB).

S: If light is incident normally on face AB, the angle of incidence on face AC is $45^{\circ }$. For TIR to occur at AC, this angle must be greater than the critical angle $C$.

So, $45^{\circ }>C\Rightarrow \sin 45^{\circ }>\sin C\Rightarrow 1/\sqrt{2}>1/\mu$.

This implies $\mu >\sqrt{2}$. The minimum refractive index for TIR is $\sqrt{2}$.