Q 9.4    Figures of (a) and (b) show refraction of a ray in air incident at 60^{\circ} with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of  incidence in water is 45^{\circ} with the normal to a water-glass interface [Fig.(c)].

        

        

Answers (1)

As we know,by snell's law

 \mu_1sin\theta _1=\mu _2sin\theta _2  where,

\mu_1 = refrective index of medium 1

\theta _1 = incident angle in medium 1 

\mu _2 = refrective index of medium 2

\theta _2 = refraction angle in medium 2

NOW, APPLYING IT FOR fig (a)   

1sin60=\mu _{glass}sin35

\mu _{glass}=\frac{sin60}{sin35}=\frac{0.866025}{0.573576 } = 1.509

Now applying for fig (b)

1sin60=\mu _{water}sin47

\mu _{water}=\frac{sin60}{sin47}=\frac{.8660}{.7313} = 1.184

Now in fig (c) Let refraction angle be \theta so,

\mu _{water}sin45=\mu _{glass}sin\theta

sin\theta =\frac{\mu _{water}*sin45}{\mu _{glass}}

sin\theta =\frac{1.184*0.707}{1.509} = 0.5546

\theta = sin^{-1}(0.5546) = 38.68

Therefore the angle of refraction when ray goes from water to glass in fig(c) is 38.68.

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