Question

Red light of 650 nm travels from air to water with an incident angle of 30 degree.The refraction index for air is 1.000293 and for water is 1.33, the speed of light in vacuum is 3x10^8 m/s. a) Find the frequency, wave length, and speed of the red light in the air; b) find the frequency, wave length, and speed of light of the red light in water; c) find the angle of the refraction

Answer 1

Answer on Question #49221-Physics-Optics

Red light of $\lambda = 650\ \mathrm{nm}$ travels from air to water with an incident angle of $\theta = 30\ \mathrm{degree}$. The refraction index for air is $n_{air} = 1.000293$ and for water is $n_{water} = 1.33$, the speed of light in vacuum is $c = 3 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}$. a) Find the frequency, wave length, and speed of the red light in the air; b) find the frequency, wave length, and speed of light of the red light in water; c) find the angle of the refraction

Solution

a) The frequency of the red light in the air

$$f_{air} = \frac{v_{air}}{\lambda_{air}} = \frac{c}{n_{air} \lambda_{air}} = \frac{3 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}}{650 \cdot 10^{-9} \mathrm{m}} = 4.62 \cdot 10^{14} \mathrm{Hz}.$$

Wave length of the red light in the air

$$\lambda_{air} = \frac{\lambda}{n_{air}} = \frac{650 \mathrm{nm}}{1.000293} = 649.8 \mathrm{nm}.$$

Speed of the red light in the air

$$v_{air} = \frac{c}{n_{air}} = \frac{3 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}}{1.000293} = 2.999 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}.$$

b) The frequency of light of the red light in water

$$f_{water} = f_{air} = 4.62 \cdot 10^{14} \mathrm{Hz}$$

Wave length of light of the red light in water

$$\lambda_{water} = \lambda_{air} \frac{n_{air}}{n_{water}} = 650 \mathrm{nm} \cdot \frac{1.000293}{1.33} = 488.9 \mathrm{nm}.$$

Speed of light of the red light in water

$$v_{water} = \frac{c}{n_{water}} = \frac{3 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}}{1.33} = 2.256 \cdot 10^{8} \frac{\mathrm{m}}{\mathrm{s}}.$$

c)

$$\frac{\sin \theta_{r}}{\sin \theta} = \frac{n_{air}}{n_{water}} \rightarrow \theta_{r} = \sin^{-1} \left(\frac{n_{air}}{n_{water}} \sin \theta\right) = \sin^{-1} \left(\frac{1.000293}{1.33} \sin 30\right) = 22{}^{\circ}.$$

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