
Refractive Indices of water and glass are $\dfrac {4} {3} $ and $\dfrac {3} {2} $ respectively. A ray of light travelling in water is incident on the water and is incident on the water glass interface at ${30^0} $. Calculate the angle of refraction.
Answer
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Hint: The refractive index of a material is a dimensionless figure that defines the rapid passage of light through the material, also known as the refraction index (or index of refraction). Refraction is an effect which takes place when a light wave moves from one medium into another, an angle away from normal, where the light velocity changes. The interface between air and glass in which it passes slower applies to light. Light is refracted. If the light speed at the interface increases, the light's wavelength must also change. As the light enters the medium, the wavelength reduces and the light wave switches direction.
Complete step by step solution:
Refractive index, known as refraction index, calculation of the bending of a light ray when it travels from medium to medium. If I is the angle incidence of the ray in the vacuum (the angle of the incoming ray to the perpendicular to the surface of a medium, known as the normal) and (r is the angle of refraction) the refractive indices (n is the ratio of the sine to the angle of refraction of the sine). In empty space separated by the speeding of its wavelength into a material, the refractive index also equals the speed of light.
Angle of incidence, $i = {30^0} $
Refractive index of water, ${n_w} = \dfrac{4}{3}$
Refractive index of water, ${n_g} = \dfrac{2}{3}$
Let the angle of refraction be $r$.
Using Snell's law of refraction:
\[{n_w}sini = {n_g} sinr\]
\[\Rightarrow \dfrac{4}{3} \times sin {30^o} = \dfrac{{3}}{2}sinr\]
$ \Rightarrow \dfrac{4}{3} \times 0.5 = \dfrac{3}{2}\sin r$
$\sin r = 0.444$
$r = {26 ^\circ} $
Note: When the light ray moves from rarer to denser the refracted ray bends towards the normal. This means that the angle of refraction in this case will be smaller than the angle of incident. However, if the ray travels from denser to rarer the refracted ray bends away from the normal. This means the angle of refraction will be greater than the angle of incident.
Complete step by step solution:
Refractive index, known as refraction index, calculation of the bending of a light ray when it travels from medium to medium. If I is the angle incidence of the ray in the vacuum (the angle of the incoming ray to the perpendicular to the surface of a medium, known as the normal) and (r is the angle of refraction) the refractive indices (n is the ratio of the sine to the angle of refraction of the sine). In empty space separated by the speeding of its wavelength into a material, the refractive index also equals the speed of light.
Angle of incidence, $i = {30^0} $
Refractive index of water, ${n_w} = \dfrac{4}{3}$
Refractive index of water, ${n_g} = \dfrac{2}{3}$
Let the angle of refraction be $r$.
Using Snell's law of refraction:
\[{n_w}sini = {n_g} sinr\]
\[\Rightarrow \dfrac{4}{3} \times sin {30^o} = \dfrac{{3}}{2}sinr\]
$ \Rightarrow \dfrac{4}{3} \times 0.5 = \dfrac{3}{2}\sin r$
$\sin r = 0.444$
$r = {26 ^\circ} $
Note: When the light ray moves from rarer to denser the refracted ray bends towards the normal. This means that the angle of refraction in this case will be smaller than the angle of incident. However, if the ray travels from denser to rarer the refracted ray bends away from the normal. This means the angle of refraction will be greater than the angle of incident.
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