306 Aptitude Test Problems in Physics
12 ha (^) n 2 sin 2 (a/2)
Therefore, if^ the
h 2 1 — n 2 sin 2 (a/2) ,
required time is
t=2 v^ /n )
= 2h / 1
v l ill — n 2 sin (^2) (a/ 2 )
- n V/ 2 — h 2 sin (a/2) n 2 sin 2 (a/2)
h (^) 1 i 1— n 2 sin 2 (a/2) )
2h
(
2
yi _n2 sin2 a' + ViS —
h (^2)
ha n sin —a )
v
12 — h^2 n^2 sin^2 (a/2)^21
If < then t =--
h 2 ..-- 1 — n 2 sin 2 (a/2) ' v
4.15. It follows from symmetry considerations
that the image of the point source S will also be
Fig. 241
at a distance b from the sphere, but on the opposite
side (Fig. 241).
4.16. An observer on the ship can see only the rays
for which sin a < 1/ngi (if sin a > thigh such
a ray undergoes total internal reflection and cannot
be seen by the observer, Fig. 242). For the angle 0,
we have the relation
n w sin 6= no sin a, sin 6= ngl sin a,
nw