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Geometrical Optics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsGeometrical Optics3 Marks
Question
A pole of length 1.00m stands half dipped in a swimming pool with water level 50.0 cm higher than the bed. The refractive index of water is 1.33 and sunlight is coming at an angle of 45° with the vertical. Find the length of the shadow of the pole on the bed.
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Answer


Shadow length $= \text{BA}' = \text{BD} + \text{A}'\text{D} = 0.5 + 0.5 \tan \text{r}$
Now, $1.33=\frac{\sin45^{\circ}}{\sin\text{r}}\Rightarrow\sin\text{r}=0.53$
$\Rightarrow\cos\text{r}=\sqrt{1-\sin^2\text{r}}=\sqrt{1-(0.53)^2}=0.85$
So, $\tan \text{r} = 0.6235$
So, shadow length = (0.5)(1 + 0.6235) = 81.2cm

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