WCCC 3

#physics#wccc#refraction

Question

Light is incident on a piece of glass. The light travels from air (assume n = 1.00) in to the glass. In the air, the light beam makes an angle of 32.0° with the normal. In the glass it makes an angle of 21.0° with the normal.

1. List the angles of incidence and refraction in this situation Angle of Incidence: 32.0°, Angle of Refraction: 21.0°
2. What is the refractive index of the glass? Snell’s Law tells us that $n_1\sin {\theta }_1=n_2\sin {\theta }_2$ ∴ Given the information from the question we can create the following equation where $n_2$ is the refractive index of the glass: $1\times \sin 32°=n_2\sin 21°$ $n_2=\frac{\sin 32°}{\sin 21°}$ ∴ The refractive index of glass is approx. 1.4787
3. The light beam re-emerges from the other side of the glass. What is the angle from the normal at which it emerges? 32.0°

Question

Calculate the speed of light in seawater that has a refractive index of 1.38

$\frac{c}{v}=n$ $\frac{3\times 10^8}{v}=1.38$ $v=\frac{3\times 10^8}{1.38}$

Question

Light travels at a speed of $2.25\times 10^{8}m{s}^{-1}$ in water and $2.29\times 10^{8}m{s}^{-1}$ in ice. If water has a refractive index of 1.33, use this information to calculate the refractive index of ice.

$n_2=\frac{c}{v}$ $n_2=\frac{3\times 10^8}{2.29\times 10^8}$ $n_2=\frac{300}{229}$ $n_2\approx 1.31$

Question

Light travels from water (n = 1.33) into glass (n = 1.60). The incident angle is 44°. Calculate the angle of refraction

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$ $1.33\sin 44°=1.60\sin {\theta }_2$ ${\sin }^{-1}(\frac{1.33\sin 44°}{1.60})={\theta }_2$ ${\theta }_2\approx 35.27$

Question

For which of the following situations can total internal reflection occur?

Incident medium	Refracting medium
air (n = 1.00)	glass (n = 1.55)
glass (n = 1.55)	air (n = 1.00)
glass (n = 1.55)	water (n = 1.33)
glass (n = 1.55)	glass (n = 1.55)

All but the first are possible