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Secondary School Physics 7 points

A ray of light travelling through air falls on the surface of a glass slab and emerges out of it parallel to incident ray. The angle of incidence I , angle of refraction in glass r and angle of emergence e . find the ratio of sini/sine and refractive index of glass in terms of sin i / sin e

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praachie26

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Answer:

According to Snell’s law, we have:

sinisinr=μgμa …(i)

By the same law, we also have:

sinr′sine=μaμg …(ii)

We assume that PQ is parallel to SR (interfaces are parallel). Since N2N1 is perpendicular to PQ and N2'N1' is perpendicular to SR, N2N1 is parallel to N2'N1'. r=r′ follows from the equality of alternate interior angles.

From equations (i) and (ii), we have:

sinr′sine=sinrsini

Taking reciprocals of both sides:

sinesinr′=sinisinr …(iii)

Since r=r′, sinr=sinr′. Multiplying both sides of (iii) by sinr, we get:

sine=sini

Since both e and i are acute, their sines being equal implies that the angles are equal, e=i. This is the result we set out to prove.

Note: that if we don’t assume that PQ is parallel to SR, our proof would not be valid, since sinesinr′=sinisinr …(iii) would not imply sine=sini. The proof would also be invalid if AE and FD are not rays in the same medium, because we would not be able to get sinr′sine=sinrsini from equations (i) and (ii).

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Answer:

According to Snell’s law, we have:

sinisinr=μgμa …(i)

By the same law, we also have:

sinr′sine=μaμg …(ii)

We assume that PQ is parallel to SR (interfaces are parallel). Since N2N1 is perpendicular to PQ and N2'N1' is perpendicular to SR, N2N1 is parallel to N2'N1'. r=r′ follows from the equality of alternate interior angles.

From equations (i) and (ii), we have:

sinr′sine=sinrsini

Taking reciprocals of both sides:

sinesinr′=sinisinr …(iii)

Since r=r′, sinr=sinr′. Multiplying both sides of (iii) by sinr, we get:

sine=sini

Since both e and i are acute, their sines being equal implies that the angles are equal, e=i. This is the result we set out to prove.

Note: that if we don’t assume that PQ is parallel to SR, our proof would not be valid, since sinesinr′=sinisinr …(iii) would not imply sine=sini. The proof would also be invalid if AE and FD are not rays in the same medium, because we would not be able to get sinr′sine=sinrsini from equations (i) and (ii).

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