What is the number on the top left

A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of

andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

Explanation:

Refraction occurs when a light wave passes through the boundary between two mediums.

When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where :

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of the 2nd medium

$\theta_1$ is the angle of incidence (the angle between the incident ray and the normal to the boundary)

$\theta_2$ is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

$n_1=1.00$ (index of refraction of air)

$n_2=1.50$ approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

$sin \theta_2 =\frac{n_1}{n_2}sin \theta_1$

And since $n_2>n_1$ , we have

$\frac{n_1}{n_2}$

And so

$\theta_2$

Which means that

The angle of incidence is greater than the angle of refraction

6 0

3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$