101
APPENDIX D
Angle of Minimum Deviation
In this appendix we are going to prove that the light’s path through the prism has to
be symmetric in order to have the minimum angle of deviation. This situation is
presented in the Figure 1 above.
According to the Snell’s Law one has
sin i1 = n sin r1 ,
n sin r2 = sin i2 ,
(D.1)
where n is the index of refraction of the prism and index of refraction of the air is set
equal to 1. Also from the geometry of the situation we have equations 5.4 and 5.5 from
the main text:
D= (i1 – r1) + (i2 – r2),
(D.2)
A = r1 + r 2 .
(D.3)
D = i1 + i2 – A
(D.4)
The minimum deviation occurs when the derivative
dD
= 0 . Using equation D.4 we have
di1
Substituting D.3 into the D.2 equation
di
dD
= 1+ 2 = 0 ,
di1
di1
(D.5)
since A has a constant value for a given prism. In order to calculate the derivative
di2
di1
involved into this equation, one shall differentiate equations D.1 and D.3, which gives
cos i1di1 = n cos r1dr1 ,
n cos r2 dr2 = cos i2 di2 ,
(D.6)
102
0 = dr1 + dr2 .
(D.7)
Dividing the second D.6 equation by the first D.6 equation gives
cos i2 di2 cos r2 dr2
=
.
cos i1di1 cos r1dr1
(D.8)
di2
= −1 ,
di1
(D.9)
dr2
= −1 .
dr1
(D.10)
Equation D.5 becomes
and equation D.7 gives
Combining equations D.8, D.9 and D.10, we obtain
cos i2 cos r2
=
.
cos i1 cos r1
(D.11)
So, we can square both sides of D.11 to get
cos 2 i2 cos 2 r2
=
.
cos 2 i1 cos 2 r1
(D.12)
Using the trigonometric identity cos2x + sin2x = 1, D.12 becomes
1 − sin 2 i2 1 − sin 2 r2
=
.
1 − sin 2 i1 1 − sin 2 r1
(D.13)
Cross multiplying:
(1 − sin i )(1 − sin r ) = (1 − sin r )(1 − sin i ) .
2
2
2
2
1
2
2
1
(D.14)
Using equation D.1
(1 − n
2
1
⎛
⎞
sin 2 r2 ) ⎜1 − 2 sin 2 i1 ⎟ = (1 − sin 2 r2 )(1 − sin 2 i1 ) ,
⎝ n
⎠
or
1
sin 2 i1 − n 2 sin 2 r2 + sin 2 r2 sin 2 i1 = 1 − sin 2 i1 − sin 2 r2 + sin 2 r2 sin 2 i1 ,
n2
1
− 2 sin 2 i1 − n 2 sin 2 r2 = − sin 2 i1 − sin 2 r2 ,
n
1 ⎞ 2
⎛
2
2
⎜1 − 2 ⎟ sin i1 − ( n − 1) sin r2 = 0,
⎝ n ⎠
1−
(D.15)
103
⎛ n2 − 1 ⎞ 2
2
2
⎜ 2 ⎟ sin i1 − ( n − 1) sin r2 = 0,
n
⎝
⎠
2
sin i1
− sin 2 r2 = 0,
2
n
sin 2 i1 = n 2 sin 2 r2 .
(D.16)
Taking the square root:
sin i1 = n sin r2 .
(D.17)
sin i1 = sin i2 .
(D.18)
So, with account of D.1
Hence, there must be symmetry:
i1 = i2 .
(D.19)