(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 45º with the normal to a water - glass interface.
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As per the given figure, for the glass - air interface:
Angle of incidence, i = 60°
Angle of refraction, r= 35°
The relative refractive index of glass with respect to air is given by Snell's law as:
$=\text{Mh}_2=6\times4.7=28.2 \ \text{cm}$
$\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$
$\frac{\sin 60^\circ}{\sin35^\circ}=\frac{0.8660}{0.5736}=1.51\dots(1)$
As per the given figure, for the air - water interface:
Angle of incidence, i = 60°
Angle of refraction, r= 47°
The relative refractive index of water with respect to air is given by Snell's law as:
$\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$
$\frac{\sin 60^\circ}{\sin47^\circ}=\frac{0.8660}{0.5736}=1.184\dots(2)$
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
$\mu^\text{w}_\text{g}=\frac{\mu^\text{a}_\text{g}}{\mu^\text{a}_\text{w}}$
$=\frac{1.51}{1.184}=1.275$
The following figure shows the situation involving the glass - water interface.
Angle of incidence, i= 45°
Angle of refraction = r
From Snell's law, rcan be calculated as:
$\frac{\sin\text{i}}{\sin\text{r}}=\mu^\text{w}_\text{g}$
$\frac{\sin45^\circ}{\sin\text{r}}=1.275$
$\sin\text{r}=\frac{\frac{1}{\sqrt2}}{1.275}=0.5546$
$\therefore \ \text{r}=\sin^{-1}(0.5546)=38.67^\circ$
Hence, the angle of refraction at the water - glass interface is 38.68°.
Angle of incidence, i = 60°
Angle of refraction, r= 35°
The relative refractive index of glass with respect to air is given by Snell's law as:
$=\text{Mh}_2=6\times4.7=28.2 \ \text{cm}$
$\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$
$\frac{\sin 60^\circ}{\sin35^\circ}=\frac{0.8660}{0.5736}=1.51\dots(1)$
As per the given figure, for the air - water interface:
Angle of incidence, i = 60°
Angle of refraction, r= 47°
The relative refractive index of water with respect to air is given by Snell's law as:
$\mu^\circ_\text{g}=\frac{\sin\text{i}}{\sin\text{r}}$
$\frac{\sin 60^\circ}{\sin47^\circ}=\frac{0.8660}{0.5736}=1.184\dots(2)$
Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:
$\mu^\text{w}_\text{g}=\frac{\mu^\text{a}_\text{g}}{\mu^\text{a}_\text{w}}$
$=\frac{1.51}{1.184}=1.275$
The following figure shows the situation involving the glass - water interface.
Angle of incidence, i= 45°
Angle of refraction = r
From Snell's law, rcan be calculated as:
$\frac{\sin\text{i}}{\sin\text{r}}=\mu^\text{w}_\text{g}$
$\frac{\sin45^\circ}{\sin\text{r}}=1.275$
$\sin\text{r}=\frac{\frac{1}{\sqrt2}}{1.275}=0.5546$
$\therefore \ \text{r}=\sin^{-1}(0.5546)=38.67^\circ$
Hence, the angle of refraction at the water - glass interface is 38.68°.
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