A puddle of water is covered with an oil film that has an index of refraction with an unknown value. However, it is known that the unknown value is greater than that of water. The thickness of the oil film is such that it is the minimum thickness possible for normally incident infrared radiation with a wavelength of 836 nm to experience destructive interference in the reflected radiation. When white light is incident normal to the oil film, what wavelengths in the visible range (400 nm-700 nm) will experience constructive interference?

Two antennai (which may or may not be in phase with each other) transmit a radio wave at an AM frequency of 1500 kHz. While driving by on a semi-circular road around these antennae, you notice that there are only three totally destructive locations (one at the centerline and two at the same angle above and below the centerline as shown). What is the path length difference of the radio waves for the topmost destructive location?

A general expression for the electric field associated with a monochromatic plane wave is E(r,t) = e.Excos(@t-k-r). Give the numerical values of Eo, w, and k (with units, and direction for vector quantities) for the case where the plane wave has a wavelength of 350 nm in vacuum, the refractive index of the medium is n = 1.46, the electric field lies in the yz plane, and the intensity is 2.0 MW/m². Use time in units of s and distance in units of m.

Sometimes a nonreflective coating is applied to a lens, such as a camera lens. The coating has an index of refraction between the index of air and the index of the lens. The coating cancels the reflections of one particular wavelength of the incident light. Usually, it cancels green-yellow light (2 = 540.0 nm) in the middle of the visible spectrum. (a) Assuming the light is incident perpendicular to the lens surface, what is the minimum thickness of the coating in terms of the wavelength of light in that coating? (Use the following as necessary: λ.) W = (b) If the coating's index of refraction is 1.37, what should be the minimum thickness of the coating? m

Consider two identical antennas separated by 9.00 m that radiate in phase at 120 MHz. A receiver placed 150 m from both antennas measures an intensity I0. The receiver is moved so that it is 1.8 m closer to one antenna than to the other. a) What is the phase difference ϕ between the two radio waves produced by this path difference? b) In terms of I0, what is the intensity measured by the receiver at its new position?

I am having trouble understanding this question, but I have a few guesses as to what the answer could be.  An ideal polarizing sheet transmits 90% of the incident partially polarized light.  The polarizing sheet is now rotated through 45 degrees; during the rotation process the transmitted light monotonically decreases in intensity. What fraction of the incident intensity now passes?   My guess is that the answer is between 0% and 50%, but I am not sure why.    Please advise, any clarification or answer would be useful! :)

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 30.0 ∘∘ and 57.0 ∘∘, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 70.0 W/cm2W/cm2 after it passes through the stack. A)If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed? Express your answer in watts per meter squared.

A beam of initially unpolarized light of intensity 227 W/m^2 passes through a series of polarizing filters, each one's axis aligned at a 11.1-degree angle with respect to the previous filter s axis. The beam of light emerging from the final filter in the series has an intensity of 64.4 W/m^2. How many polarizing filters did the beam pass through? 24 16 8 32

I am having difficulty understanding polarization of sheets and the incident intensity of light that passes through it. My question has three parts: PART A: If a vertically oriented, ideal polarizing sheet transmits 50% of the incident unpolarized light and the polarizing sheet is now rotated through 45 degrees, what fraction of the incident intensity now passes?  My guess is that 0% of the incident intenisty would pass through, but I am not sure if I am wrong and 100% or 50% of the incident intensity passes through instead. Any answer or explaination would be greatly appreciated.    PART B: Now, if an ideal polarizing sheet transmits 90% of the incident partially polarized light, what percentage of the incident light is polarized? Here, I am assuming 100% of the light is polarized, but I am not entirely sure. Any answer or explaination would be greatly appreciated.    PART C: Similar to PART B, an ideal polarizing sheet transmits 90% of the incident partially polarized light. But…

When light passes through sequential interfaces, there may be a change of phase upon reflection at each interface. Additionally the wavelength changes with each change of the index of refraction. If the thickness of a thin layer is chosen correctly, then it will serve as an anti-reflective coating bringing about maximally destructive interference between the first and second reflected waves.For convenience, the index of refraction for a variety of materials is provided below. Materials at 20°C for light with a vacuum wavelength of 589 nm. Material n   Material n   Material n benzene 1.501   diamond 2.419   plexiglas 1.51 carbon disulfide 1.628   fluorite (CaF2) 1.434   quartz (crystalline) 1.544 carbon tetrachloride 1.461   glass (crown) 1.52   quartz (fused) 1.458 ethanol 1.361   glass (flint) 1.66   sodium chloride 1.544 glycerine 1.473   ice (0°C) 1.309   zircon 1.923 water (fresh) 1.333   polystyrene 1.49   air Given that the thin layer of diamond has thickness…