760
15 Optics
15.64 r n =
f n nλ
r 1 =
50 × 1 × 5 × 10 −5 = 0.05 cm
r 25 =
50 × 25 × 5 × 10 −5 = 0.25 cm
15.3.6 Polarization
15.65 Path difference introduced by the plate
x = (n 0 − n e )t = (1.642 − 1.478) × 4 × 10
−5
= 6.56 × 10
−6 m
Phase difference
δ =
2ππx
λ
=
(2π)(6.56 × 10 −6 )
6 × 10 −7
= 68.66 rad
15.66 I θ = I max cos 2 θ
(Malus’ law)
cos θ =
I θ
I max
=
1
2
=
1
√
2
∴ θ = ±45, ±135
◦
.
15.67 tan θ p = μ = 1.33
(Brewster’s law)
∴
θ p = 53
◦
Therefore the elevation of the sun is 90 − 53 ◦ = 37 ◦
15.68 tan Q p = μ wg =
μ g
μ w
=
1.5
1.33
= 1.1278
∴ Q p = 48.4 ◦
15.69 t =
λ
4(μ 0 − μ e )
=
6 × 10 −5
4(1.553 − 1.544)
= 1.67 × 10 −3 cm
15.70 θ = αL D
where θ is the angle of rotation of plane of polarization, α is the specific
rotation, L is the length of the tube in decimetres and D is the amount of
solvent in grams per 100 c.c:
L = 20 cm = 2 dm