# 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 = 26.1

^{o}

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 90

^{o}it passes into the block undeflected.

At the plastic /water interface the angle of incidence = 45

^{o}

n

_{1}Sin q1 = n

_{2}Sin q2

1.62 Sin 45 = 1.33 Sin q2

q2 = 59.5

^{o}

At the water/ air interface: the angle of incidence = 180 - 135 - 30.5 = 14.5

^{o}

n

_{1}Sin q

_{1}= n

_{2}Sin q

_{2}

1.33 Sin 14.5 = 1 Sin q

_{2}

q

_{2}= 19.5

^{o}.

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