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'O' Level Standard Questions - Refraction

Q10.

A rectangular glass block 12 cm thick is placed over an ink mark on a sheet of paper. Find the distance the ink mark is apparently raised if the refractive index of the glass is 1.50.

 

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 leaving the glass, but the ones around them will. They will bend away from the normal as they exit the glass.

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 position nearer the surface of the block - the apparent depth. This makes the ink appear to be within the block not under it!

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= nglasssin θglass

 

1 x sin β = nglassx sin α

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

1.50 = 12/apparent depth

apparent depth = 12/1.5 = 8.0 cm = distance xy

The distance the ink has apparently risen = distance xz - distance xy

= 12 - 8.0 = 4.0

The ink is therefore apparently raised 4.0 cm into the block