Candidates are expected to have a thorough understanding of the syllabus details outlined in the accompanying figure. Refraction of light. Light will refract when it travels through the transparent materials, such as glass. Angle of incidence is denoted by I. Angle of refraction is denoted by R.
As the incident light ray enters the glass block, it slows down and bends towards the normal. As the light ray leaves the glass block, it speeds up and bends away from the normal. The frequency of light is unchanged as it travels from one medium to another. The wavelength of light changes directly with the speed of light. As the light speed up, the wavelength increases.
As the light slows down, the wavelength decreases. When light travels from an optically less dense medium to a more dense medium, it slows down and so bends towards the normal. When light travels from a more dense to less dense medium, it speeds up and bends away from the normal.
If the angle of incidence is zero degrees, the ray enters along the normal to the surface of the glass block. The light slows down but does not change direction because the angle of refraction is also zero. As it leaves it speeds up.
but does not change direction. Refractive index. Refractive index, n, of a material, is the ratio of speed of light in a vacuum, c, to speed of light in the material, v. We can write the equation of refractive index as n equals c over v, where n is a refraction index, which is a constant for the medium and has no unit. C is the speed of light in vacuum, which is equal to 3 times 10 to the power of 8. V is the speed of light in a medium.
The refraction index in air is approximately equal to 1, because speed of light in air is very close to the speed of light in vacuum. The refractive index of a material is related to how dense it is. Generally, the denser the material, the higher the refractive index.
For example, The light ray in the medium A bends less than the light ray in the medium B. So, the speed of light in the medium A is more than the speed of light in the medium B. This shows that the medium A is less dense than the medium B. Therefore, the refractive index in the medium A is less than in the medium B. Snell's law.
Snell's law is a formula used to describe the relationship between the angles of incidence and refraction. N1 sine I equals N2 sine R. Where N1 is the refractive index in the first medium that light passes, I is the angle of incidence.
N2 is the refractive index in the second medium that light passes. R is the angle of refraction. When light travels from air to a medium, N1 is refractive index in air, which is equal 1. So the formula becomes n equals sine i over sine r. Example 1. A ray of light is incident in air on a glass at an angle of 60 degrees to the normal.
The refractive index of glass is 1.5. Calculate the angle of refraction. From Snell Law.
Substitute n. 1 equals 1 for air. I equals 60. An n 2 equals 1.5 for glass.
So sine r equals sine 60 over 1.5. Then calculate r to get r equals 35.3 degrees. Example 2. A ray of light in water incident to the air at the refractive angle of 70 degrees. The refractive index of water is 1.3.
Calculate the angle of incidence. From Snell's law. Substitute, n1 equals 1.3 for water.
n2 equals 1 for air. nr equals 70 degrees. So, sine i equals sine 70 over 1.3.
Then calculate R to get R equals 46.3 degrees. Example 3. Array of light in air incident to the diamond at the incident angle of 47 degrees and the refractive angle of 15 degrees. Calculate the refractive index and the speed of light in the diamond.
From Snell's law. Substitute N1 equals 1 for air. I equals 47 degrees. An R equals 15 degrees. So, N equals sine, 47, over sine, 15. N equals 2.8.
From N equals C over V. So, V equals C over N. Substitute, N equals 2.8. And C equals 3 times 10 to the power of 8. Then, V equals 1.07 times 10 to the power of 8. 8 meters per second. An experiment to investigate the reflection of light and calculate the speed of light in the glass. Place a glass block on a piece of paper and trace around the block with a pencil.
Shine a ray of light in air on the glass block. Place two pins along the incident ray with a separation of more than 5 centimeters to ensure accuracy when drawing the line. Place two pins along the emerging ray from other side of glass block, with a separation of more than 5 cm.
Remove the glass block and pins, then mark at the pins'holes. Turn off the light box. Use a ruler to draw the incident ray and the emerging ray. Then, connect the entry and exit points to show the path of light within the glass block. Draw the normal line at the point of incidence.
Measure the angle of incidence, I, and the angle of refraction, R. Calculate the refractive index of the glass using n equals sine, I, over sine, R. Calculate the speed of light in the glass using the formula V equals C over n.
Repeat the experiment for five different angles of incidence, and then find the average of refractive index and speed of light in the glass. Or, plot the graph of sine, I. against sine r. Then find the refractive index by the gradient of graph. Calculate the speed of light in the glass using v equals c over n.
The critical angle and the total internal reflection in a medium. As the ray shine on the semicircle glass block, the light slows down but does not change direction. This is because the angle of incidence is zero degrees.
the ray enters along the normal to the surface of the glass block, then the angle of refraction is also zero. As the ray reach the point A, some ray reflect and some ray refract. When a ray of light travels from a denser medium to a less dense medium, it bends away from the normal. As the angle of incidence increases, the angle of refraction increases. The intensity of refractive ray decreases, while intensity of reflective ray increases.
When r equals 90 degrees, the angle of incidence is equal to the critical angle c. Therefore, the critical angle is the is the angle of incidence in denser medium, where the angle of refraction is 90 degrees. From Snell's law, substitute n1 equals n, which is the refractive index in the medium.
I equals C, which is the critical angle. N2 equals 1, which is the relative index in air. R equals 90 degrees.
So, the result is N equals 1 over sine C. If the angle of incidence is increased beyond the critical angle, total internal reflection occurs. The total internal reflection occurs when the light ray travels from denser medium to less dense medium.
The angle of incidence is more than the critical angle of the medium. No refraction of light. The angle of incidence is equal to the angle of reflection. Uses of the total internal reflection in everyday life.
Periscope consists of the glass prisms, which are used in submarines. As a light ray travels through a periscope, Total internal reflection occurs at point A and point B. A glass right-angle prism with an angle 45 degrees.
The critical angle of glass is about to 43 degrees. As the light ray travels from air to glass at the angle of incidence is zero, so the light slows down, but does not change direction. The light ray reflects at the point A with the incident angle of 45 degrees, which is greater than the critical angle of glass, 43 degrees.
This causes the light ray to undergo total internal reflection. As it leaves the prism, it speeds up but does not change direction. Binocular consists of the prisms as shown.
As the light ray travels through a binocular and the total internal reflection occurs at points A, B, C and D. The prisms in this binocular are glass right angle prism 45 degrees. As the light ray travels from air to glass at the angle of incidence of zero, so the light slows down but does not change direction. The light ray reflect at the point A with the angle of incidence is 45 degrees, which is greater than the critical angle of glass. This causes the total internal reflection to occur.
Then the light ray reflect at the point B with the angle of incidence is 45 degrees, which is greater the critical angle of glass. This causes the total internal reflection to occur again. As it leaves from the prism, it speeds up.
but does not change direction. Rear Reflector. Rear reflectors use total internal reflection to reflect light back to the source. They are often used on vehicles to improve their visibility to other road users.
Optical Fibers. An optical fiber is a thin glass cylindrical core coated with a transparent material of lower refractive index cladding. The cladding has a lower refractive index than the core, so it is less dense. This means that total internal reflection will occur for all rays of light that strike the boundary between the core and cladding at an angle greater than the critical angle. Optical fibers can be used to communicate signals, for example telephone conversations, or in medical applications, such as endoscopy, a method of examining inside the body.
In the communications, optical fibers are used in the TV, internet, and phones. Light or infrared carries the information along the optical fibers. They can carry a large amount of information at very high speed.
In medicine, optical fibers are used in the endoscopes that allow the surgeons to see inside the body of their patients. Light ray from the outside is transmitted through the optical fibers and undergo total internal reflection, which prevents them from leaving the optical fibers. Light ray reflected at the inside the body and return back to the outside along the other optical fibers. This light is sent to the computer monitor. I hope you found this video helpful.
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