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'O' Level Standard Questions - Refraction

Q3. 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?

Snell's Law: n1sin θ1 = n2sin θ2

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