15.3 Solutions
727
A function f (x) is said to have a stationary value at x = x 0 , if its derivation
d f /dx vanishes at x = x 0 . A stationary value could correspond to a maximum
or minimum.
Let a ray of light proceed from A(0, y 1 ) in medium of index n 1 and be incident
at C(x, 0) on the interface, get refracted and reach B(x 2 , y 2 ) in the medium of
index n 2 :
O.P.L. = n 1 (AC) + n 2 (CB) = n 1
x 2 + y 2
1 + n 2
(x 2 − x) 2 + y 2
2
d(O.P.L)
dx
= 0
∴
n 1 x
x 2 + y 2
1
−
n 2 (x 2 − x 1 )
(x 2 − x) 2 + y 2
2
= 0
∴ n 1 sin i − n 2 sin r = 0
∴
sin i
sin r
=
n 2
n 1
(Snell’s law)
(3)
15.4 A mirage is a type of illusion formed by light rays coming from the low region
of the sky in front of the observer. On a sunny day, a road gets heated and a
temperature gradient is established in the vertical direction with the upper air
layer being slightly less warmer and the corresponding indices of refraction
being slightly larger. As the rays penetrate the atmospheric depth they start
bending due to refraction, becoming horizontal to the road and then bending
upwards. The blue of the sky in the background produces a virtual image of a
water pool, and the turbulence of air close to the road enhances the effect of a
water pool with waves on the surface. This phenomenon is known as mirage.
(a) Referring to Fig. 15.5, optical path length (O.P.L) =
2
i=1
n i s i = n 1 AB +
n 2 BC + n 1 CD
= n 1
(d − x) 2 /4 + h 2 + n 2 x + n 1
(d − x) 2 /4 + h 2
O.P.L = n 2 x + n 1
(d − x) 2 + 4h 2
(1)
(b)
d(O.P.L)
dx
= 0
(Fermat’s principle)
Using (1), n 2 −
n 1 (d − x)
(d − x) 2 + 4h 2
= 0
Substituting n 1 = 1.00030, n 2 = 1.00020 and h = 2 m and solving
d − x = 282.8
( 2 )
or
x = 500 − 282.8 = 217.2 m