A small piece of wood floats on the surface of a 2.5 m deep lake. Determine the position where the shadow forms on the bottom when the sun is just setting. The refractive index of water is 4/3 .

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Quantrex Academy · Accepted Answer

Correct answer: C. 2.83 m shifted from the position directly below the piece of wood

When the sun is just setting, the angle of incidence of the light rays is

i = 90^{{}^{{}^{\circ}}}

.

Using Snell's law:

\sin(90^{{}^{{}^{\circ}}}) = \mu \sin(r)

1 = \dfrac{4}{3} \sin(r) \Rightarrow \sin(r) = \dfrac{3}{4}

The tangent of the angle of refraction is:

\tan(r) = \dfrac{\sin(r)}{\sqrt{1 - \sin^2(r)}} = \dfrac{3/4}{\sqrt{1 - (3/4)^2}} = \dfrac{3}{\sqrt{7}}

The horizontal shift of the shadow from the position directly below the piece of wood is:

x = h \tan(r)

Substituting

h = 2.5 \text{ m}

:

x = 2.5 \times \dfrac{3}{\sqrt{7}} = \dfrac{7.5}{\sqrt{7}}

Using

\sqrt{7} \approx 2.645

:

x \approx \dfrac{7.5}{2.645} \approx 2.83 \text{ m}

A small piece of wood floats on the surface of a 2.5 m deep lake....

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