Chapter 18 - Worked Solutions

to selected extension questions from the OUP text New Century Senior Physics -
by Walding, Rapkins and Rossiter

Q39.

Need to find the value of x or PN, Angle of incidence = 41
Sin i / Sin r = n
Sin r = Sin i / n = sin 41 / 1.49 or r = 26.1o
Cos r = 5 / MP
MP = 5 / Cos 26.1
MP = 5.57 cm
In triangle MNP, Sin (41 - 26.1 ) = PN / MP
PN = MP Sin 14.9 = 5.5 Sin 14.9 = 1.43 cm
Q 40.

Since the incident ray on the plastic block is 90o it passes into the block undeflected.
At the plastic /water interface the angle of incidence = 45o
n1 Sin q1 = n2 Sin q2
1.62 Sin 45 = 1.33 Sin q2
q2 = 59.5o
At the water/ air interface: the angle of incidence = 180 - 135 - 30.5 = 14.5o
n1 Sin q1 = n2 Sin q2
1.33 Sin 14.5 = 1 Sin q2
q2 = 19.5o.
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