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(C) The ray from the bottom centre $C$ travels to the top edge and then refracts into the air towards the observer.Using Snell's law at the water-air

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$\frac{8}{3 \sqrt{53}}$

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(C) The ray from the bottom centre $C$ travels to the top edge and then refracts into the air towards the observer.Using Snell's law at the water-air interface: $n_1 \cdot \sin i = n_2 \cdot \sin r$.Here,the angle of incidence $i$ in water is the angle the ray makes with the normal. Since the ray makes an angle $	heta$ with the horizontal,the angle with the vertical normal is $i = 90^{\circ} - 	heta$.The angle of refraction $r$ is the angle the ray makes with the normal in air. From the geometry,$	an r = \frac{x/2}{h} = \frac{x}{2h}$.Given $\frac{h}{x} = \frac{7}{4}$,we have $	an r = \frac{1}{2(h/x)} = \frac{1}{2(7/4)} = \frac{1}{3.5} = \frac{2}{7}$.Thus,$\sin r = \frac{2}{\sqrt{2^2 + 7^2}} = \frac{2}{\sqrt{4 + 49}} = \frac{2}{\sqrt{53}}$.Applying Snell's law: $n_{water} \cdot \sin i = n_{air} \cdot \sin r$.$\frac{4}{3} \cdot \sin(90^{\circ} - 	heta) = 1 \cdot \sin r$.$\frac{4}{3} \cdot \cos 	heta = \frac{2}{\sqrt{53}}$.$\cos 	heta = \frac{2}{\sqrt{53}} \cdot \frac{3}{4} = \frac{6}{4\sqrt{53}} = \frac{3}{2\sqrt{53}}$.Wait,re-evaluating the geometry: The ray from $C$ to the edge makes an angle $r$ with the vertical. $	an r = \frac{x/2}{h} = \frac{x}{2h} = \frac{1}{2(7/4)} = \frac{2}{7}$. Correct.Snell's law: $1 \cdot \sin(90^{\circ}-	heta) = \frac{4}{3} \sin r \Rightarrow \cos 	heta = \frac{4}{3} \cdot \frac{2}{\sqrt{53}} = \frac{8}{3\sqrt{53}}$.

$A$ student sees the top edge and the bottom centre $C$ of a pool simultaneously from an angle $\theta$ above the horizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is $\frac{4}{3}$. If $\frac{h}{x}=\frac{7}{4}$,then $\cos \theta$ is

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