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

Answers (1)

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As we know the snell's law :

$\mu_1sin\theta _1=\mu _2sin\theta _2$

where,

$\mu_1$ = refractive index of medium 1

$\theta _1$ = incident angle in medium 1

$\mu _2$ = refractive index of medium 2

$\theta _2$ = refraction angle in medium 2

Now applying it to Fig. (a), we get

$1sin60=\mu _{glass}sin35$

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

Now applying it to Fig (b), we get

$1sin60=\mu _{water}sin47$

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

Now in Fig (c), Let the refraction angle be $\theta$ so,

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

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

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

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

Hence, the angle of refraction when the ray goes from water to glass in Fig(c) is 38.68.

Posted by

Pankaj Sanodiya

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