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Lambert’s law of absorption states that the absorption of light in a thin transparent layer is...
A thin layer of air is between 2 pieces of glass. When white light is incident upon them, green light of 520 nm is reflected back strongly, but blue light of 433.33 nm appears dark. What is a possible thickness of the air layer?
Transmission through thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and r4 (the light reflects twice inside material 2). The waves of r3 and r4 interfere, and here we consider the type of interference to be either...
I I know that the thickness = lambda/(2nsin 30) BUT why? I've also attached some notes I have! (Please explain using diagram and words for me to rate the answer) We were unable to transcribe this imageThin-Film Interference When light is incident on a thin transparent film, the light waves reflected from the front and back surfaces interfere. For near-normal incidence, the wavelength con ditions for maximum and minimum intensity of the light reflected from a film in air are...
This question uses the inverse-square law, which states that the light intensity experienced at a distance x from a light source is proportional to the luminosity of the source times x-2. It is the dead of night in Gormenghast Castle, and Steerpike, on the run, is trying to hide in a long, narrow corridor that is dimly lit by two lights, one at each end. (a) Steerpike knows that his best chance of hiding is to be in the spot...
2. Newton's law of cooling can be used to derive a differential equation model for the temperature of an object that is placed in a cool medium. We define the variables u(t) to represent the temperature of the object at time t. We also define the parameter T to be the constant temperature of the medium. Newton's law of cooling states that the rate of change in the temperature of the object is proportional to the difference between the temperature...
To understand polarization of light and how to use Malus's law to calculate the intensity of a beam of light after passing through one or more polarizing filters. The two transverse waves shown in the figure(Figure 1) both travel in the +z direction. The waves differ in that the top wave oscillates horizontally and the bottom wave oscillates vertically. The direction of oscillation of a wave is called the polarization of the wave. The upper wave is described as polarized...
A planar, monochromatic light wave falls normally onto a glass plate (n =1.5), which is fabricated to have two regions of different thicknesses (see Figure 1, right panel). 2/3, where A is the wavelength of the incident light. Light passing through the plate is collected by a lens. Find the light intensity I in the focus of the lens, if the intensity collected with a plate having d 0 is Io. (2p) The thickness difference d A thin lens made...
law: I = I0 cos²θ where I0 is the intensity of the polarized light beam just before entering the polarizer, I is the intensity of the transmitted light beam immediately after passing through the polarizer, and is the angular difference between the polarization angle of the incident beam and the transmission axis of the polarizer. After passing through the polarizer, the transmitted light is polarized in the direction of the transmission axis of the polarizing filter. Part DOne way to produce a beam of polarized...
(17 %) Problem 3: A thin layer of oil with index of refraction no-l .47 is floating above the water. The index of refraction of water is nw 13. The index of refraction of air is na-1. A light with wavelength λ = 625 nm goes in from the air to oil and water. 25% Part (a) Express the wavelength of the light in the oil, λο, in terms of λ and no. 쇼 25% Part (b) Express the minimum...
Gamma-rays (Y-rays) are highly energetic electromagnetic radiation that are emitted when the nuclei undergo transitions in the energy levels. The energy of the y-rays is upward of 100 keV. Due to the high energy, the rays are not appreciably absorbed by even several centimeters of materials such as concrete or wood. However, dense metals such lead, are effective Y-ray absorbers. The amount y-rays that are absorbed depends on the material and the energy of the gamma-rays, the higher the energy,...