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Refraction of ray passing through tank filled with water An observer is viewing along the line shown in Fig. 34-4. When there is no water filled in the tank, he can see only wall AB and no other part of the base of the tank. Then water is filled in till height h and he is just able to see the point E as shown in Fig. 34-3. Find the angle of refraction into air (phi) and angle of incidence from water (phi). Also find the depth to which water has been filled. |
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Answer» Solution :Here, we can SEE that refraction of ray enables the observer to see E. So, we can apply Snell.s law and geometry. Calculation : From Fig. 34-3 , we can see that the normal and wall AB are parallel to each other. So, angle of refraction into air is `tantheta=((160)/(3))/(40)impliestheta=53^(@)` From Snell.s law, we can say that `1xxsin theta=(4)/(3)sinphiimpliesphi=37^(@)` From Fig. 34-3, we can say that `tan53^(@)=(x)/(40-h)` and `tan37^(@)=(30-x)/(h)`. Solving, we get `h=40cm`.  Figure 34-3 when the water is filled in a TANK, we can see the hidden part of the tank as well with the help of refracted RAYS. |
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