Figures (a) and (b) show refraction of a ray in air incident at 60° 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° with the normal to a water-glass interface. [ Fig. 9.34 (c)].

Solution

(a)
Given,

$Angle of incidence, i = 60 ° Angle of refraction, r = 35 °$

Using Snell's law, we have

Refractive index of glass with respect to air, ${}^{a}μ_{g} = \frac{\sin i}{\sin r} = \frac{\sin 60 °}{\sin 35 °}$

$= \frac{0.8660}{0.5736} = 1.51$

(b)

$Angle of incidence, i = 60 ° Angle of refraction, r = 41 °$

Using Snell's law,

Refractive index of water wr.to air, ${}^{a}μ_{w} = \frac{\sin i}{\sin r} = \frac{\sin 60 °}{\sin 41 °} = 1.32$

(c)
Angle of incidence in water, $i = 45 °$

Refractive index of glass wr.to water,
${}^{ω}μ_{g} = \frac{{}^{a}μ_{g}}{{}^{a}μ_{w}} = \frac{\sin i}{\sin r}$
$\frac{1.51}{1.32} = \frac{\sin 45 °}{\sin r} = \frac{0.7071}{\sin r}$

$∴ \sin r = \frac{1.32 \times 0.7071}{1.51} = 0.6181$

$\Rightarrow$ $r = 38.2 °, w h i c h i s t h e r e q u i r e d a n g l e o f r e f r a c t i o n o f g l a s s .$

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Wave Optics

Figures (a) and (b) show refraction of a ray in air incident at 60° 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° with the normal to a water-glass interface. [ Fig. 9.34 (c)].

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