The hottest would be the O type and the coolest is M
The three properties of electromagnetic waves are; they travel at the speed of light, they include ultraviolet waves, and they can transfer energy through empty space.
<h2>Further Explanation</h2><h3>A wave</h3>
- A wave is a transmission of a disturbance. It involves transmission of energy from one point which is the source to another point.
- Waves may be classified depending on the need for a transmission medium or based on the vibration of particles relative to the direction of wave motion.
- Waves may be either transverse or longitudinal based on the direction of wave motion relative to the vibration of particles
- Additionally waves may be classified as either electromagnetic wave or mechanical based on the need for a transmission medium.
<h3>Electromagnetic waves </h3>
- Electromagnetic waves are types of waves that do not require a material medium for transmission.
- All waves of the electromagnetic spectrum are electromagnetic transverse waves that do not require a material medium for transmission.
- They include; radio waves, microwaves, infrared, visible light, ultra-violet, x-rays, and gamma rays.
- All waves of the electromagnetic spectrum travel with a speed of light, 3.0 x10^8 m/s.
- Additionally, electromagnetic waves possess energy that is given by; E = hf; where h is the plank's constant and f is the frequency.
keywords: Wave, electromagnetic wave, electromagnetic spectrum
<h2>Learn more about: </h2>
Level: High school
Subject: Physics
Topic: Electromagnetic spectrum
Sub-topic: Properties of an electromagnetic waves
A persons or animals nature, especially as it permanently affects their behavior
Dispersion angle = 0.3875 degrees.
Width at bottom of block = 0.09297 cm
Thickness of rainbow = 0.07038 cm
Snell's law provides the formula that describes the refraction of light. It is:
n1*sin(θ1) = n2*sin(θ2)
where
n1, n2 = indexes of refraction for the different mediums
θ1, θ2 = angle of incident rays as measured from the normal to the surface.
Solving for θ2, we get
n1*sin(θ1) = n2*sin(θ2)
n1*sin(θ1)/n2 = sin(θ2)
asin(n1*sin(θ1)/n2) = θ2
The index of refraction for air is 1.00029, So let's first calculate the angles of the red and violet rays.
Red:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.641) = θ2
asin(1.00029*0.653420604/1.641) = θ2
asin(0.398299876) = θ2
23.47193844 = θ2
Violet:
asin(n1*sin(θ1)/n2) = θ2
asin(1.00029*sin(40.80)/1.667) = θ2
asin(1.00029*0.653420604/1.667) = θ2
asin(0.39208764) = θ2
23.08446098 = θ2
So the dispersion angle is:
23.47193844 - 23.08446098 = 0.38747746 degrees.
Now to determine the width of the beam at the bottom of the glass block, we need to calculate the difference in the length of the opposite side of two right triangles. Both triangles will have a height of 11.6 cm and one of them will have an angle of 23.47193844 degrees, while the other will have an angle of 23.08446098 degrees. The idea trig function to use will be tangent, where
tan(θ) = X/11.6
11.6*tan(θ) = X
So for Red:
11.6*tan(θ) = X
11.6*tan(23.47193844) = X
11.6*0.434230136 = X
5.037069579 = X
And violet:
11.6*tan(θ) = X
11.6*tan(23.08446098) = X
11.6*0.426215635 = X
4.944101361 = X
So the width as measured from the bottom of the block is: 5.037069579 cm - 4.944101361 cm = 0.092968218 cm
The actual width of the beam after it exits the flint glass block will be thinner. The beam will exit at an angle of 40.80 degrees and we need to calculate the length of the sides of a 40.80/49.20/90 right triangle. If you draw the beams, you'll realize that:
cos(θ) = X/0.092968218
0.092968218*cos(θ) = X
0.092968218*cos(40.80) = X
0.092968218*0.756995056 = X
0.070376481 = X
So the distance between the red and violet rays is 0.07038 cm.