A 1.00 m long pole stands half-dipped in a swimming pool where the...

The total length of the pole is $100 \text{ cm}$. Since it is half-dipped in the pool, the length of the pole above the water surface is $50 \text{ cm}$ and the length below the water surface is $50 \text{ cm}$.

The sunlight strikes at an angle of 45^{}^{{}^{\circ}} with the vertical. The shadow cast by the portion of the pole above the water on the water surface has a length $x_1$ given by:

x_1 = 50 \tan 45^{}^{{}^{\circ}} = 50 \text{ cm}

The light ray from the tip of the pole strikes the water surface at an angle of incidence i = 45^{}^{{}^{\circ}}. Let the angle of refraction be $r$. Applying Snell's law at the air-water interface:

$\sin i = \mu \sin r$

\sin 45^{}^{{}^{\circ}} = 1.33 \sin r

$\sin r = \dfrac{1}{1.33\sqrt{2}}$

Taking $\sqrt{2} \approx 1.41$, we get:

$\sin r \approx \dfrac{1}{1.33 \times 1.41} \approx 0.533$

The tangent of the angle of refraction is:

$\tan r = \dfrac{\sin r}{\sqrt{1 - \sin^2 r}} \approx \dfrac{0.533}{\sqrt{1 - (0.533)^2}} \approx 0.630$

The refracted ray travels a vertical depth of $50 \text{ cm}$ to reach the bed of the pool. The horizontal distance $x_2$ covered by the refracted ray in the water is:

$x_2 = 50 \tan r = 50 \times 0.630 = 31.5 \text{ cm}$

The total length of the pole's shadow on the bed is the sum of the horizontal distances:

$L = x_1 + x_2 = 50 + 31.5 = 81.5 \text{ cm}$

Answer: $81.5 \text{ cm}$