'O' Level Standard Questions - RefractionQ3. A diver is working under water and a ray of light from his lamp strikes the surface of the water at an angle of 55° to the horizontal. Take the refractive index of the water as being 1.33.
At what angle to the horizontal will the ray travel after it leaves the water?
We know that nair is 1 - as the refractive index is the ratio of the velocity of light in a vacuum compared to its velocity in the medium - and the speed of light in air is virtually the same as in a vacuum. Therefore if we work to about 3sf there is no difference. We also know that angles are measured from the normal - therefore in this case we need to work out the angle between the beam and a line at rightangles to the surface of the water. 90o - 55o = 35o nairsin θair = nwatersin θwater sin θwater = 1.33 x sin 35o sin θwater= 0.762 θwater= 50o This is the angle that the refracted ray makes with the normal to the water surface. Therefore the angle the beam makes with the water surface will be 90o - 50o = 40o
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