A glass plate 2.50 mm thick, with an index of refraction of...

Name: Problem 43, Chapter 33: University Physics with Modern Physics
Uploaded: 2019-08-06T15:44:39-08:00
Duration: 4 min 8 s
Channel: Salamat Ali
Description: A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen?

A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen?

Key Concepts

Wavelength

Wavelength is the distance over which the wave's shape repeats and is a fundamental property of light. In vacuum, it is defined by the light's frequency and the speed of light, but when light travels through a medium, its wavelength is reduced while its frequency remains constant.

Refractive Index

The refractive index is a measure of how much light slows down in a medium compared to vacuum. It affects the wavelength of light in the medium and is used to determine the optical properties such as refraction, reflection, and the phase of the light as it travels through different materials.

Optical Path Length

The optical path length (OPL) is the product of the physical distance that light travels in a medium and the medium's refractive index. It represents the effective distance that light would have traveled in vacuum and is a key concept when determining how many wavelengths fit into a given path.

Transcript

00:01 The problem is following.

00:02 We have a light source.

00:04 The light source have got a wavelength of 540 nanometers and we have a glass which has got a width of 2 .5 mm.

00:23 And that is placed between screen and the source.

00:32 So we have a glass between source and screen.

00:35 So here we have a screen.

00:39 The distance between a source and screen is 1 .8 centimeter.

00:47 The question is asked that how many wavelengths do we have in the distance of 1 .8 centimeters from a light source to the screen? how many wavelengths we have? so in order to find a wavelength, the number of a wavelength by definition, the number of wavelengths is equal to the distance the distance of a material represented by a d divided by the wavelength of a light inside that material lambda so we can write the total distance for air and a glass and divided by the the wavelength inside that material and then we will get the total number of wavelengths.

01:44 So the total number of wavelengths we have here, can be it in a distance in air, distance in air, over the wavelength plus the distance in a glass times refractory index of a glass n over the wavelength lambda in a material.

02:10 We can write a lambda in a material by this relation.

02:14 Lambda is equal to in a material is equal to lambda in a vacuum over a refractor index.

02:24 Then we can find the distance in here also by subtracting the distance of the glass from the total distance.

02:34 So the total distance we have is 0 .018 meter minus the distance of the glass...

Please give Ace some feedback