Topic Menu

Cyber-chess a chess instruction site

'O' Level Standard Questions - Refraction

Q9.

A stone lies at the bottom of a lake 8.0 m deep. To an observer directly above the stone it only appears to be 6.0 m below the surface. What is the refractive index of water?

When you look vertically down on an object you take in rays of light coming normally and nearly normally from the surface of the object.

Those exactly normal to the surface will not change direction when leaing the water, but the ones around them will. They will bend away from the normal as they exit the water.

Your eye recieves these rays and your brain extropolates them back to a point of origin. This is where you 'see' the object as being. as the rays have bent away from the normal they appear to be coming from a shallower position - the apparent depth.

Consider triangle VXZ

Sin α = VX/real depth

Consider triangle VXY

Sin β = VX/apparent depth

Therefore

VX = (Sin α) x (real depth) = (Sin β) x (apparent depth)

real depth/apparent depth = sin β/sinα

Snell's Law: nairsin θair= nwatersin θwater

 

1 x sin β = nwaterx sin α

∴ nwater= sin β/sinα = real depth/apparent depth

nwater= 8.0/6.0 = 1.3