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To produce a minimum reflection of wavel...

To produce a minimum reflection of wavelength near the middle of visible spectrum (550 nm), how thick should a coating of `MgF_(2) (mu = 1.38)` be vaccum-coated on a glass surface?

A

`10^(-7)` m

B

`10^(-10)` m

C

`10^(-9)` m

D

`10^(-8)` m

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The correct Answer is:
To solve the problem of determining the thickness of the MgF₂ coating needed to produce minimum reflection for a wavelength of 550 nm, we can follow these steps: ### Step 1: Understand the condition for minimum reflection For a thin film to produce minimum reflection, the path difference between the light waves reflected from the top and bottom surfaces of the film must be an odd multiple of half the wavelength. This condition can be expressed as: \[ \Delta x = (2n - 1) \frac{\lambda}{2} \] where \( n \) is an integer (1, 2, 3, ...) and \( \lambda \) is the wavelength of light in vacuum. ### Step 2: Calculate the total path difference The total path difference for light reflecting off the two surfaces of the film is given by: \[ \Delta x = 2 \mu t \] where \( \mu \) is the refractive index of the film (MgF₂ in this case), and \( t \) is the thickness of the film. ### Step 3: Set the two expressions for path difference equal To find the thickness \( t \) that results in minimum reflection, we set the two expressions for path difference equal: \[ 2 \mu t = (2n - 1) \frac{\lambda}{2} \] ### Step 4: Solve for thickness \( t \) Rearranging the equation to solve for \( t \): \[ t = \frac{(2n - 1) \lambda}{4 \mu} \] ### Step 5: Substitute known values We are given: - Wavelength \( \lambda = 550 \, \text{nm} = 550 \times 10^{-9} \, \text{m} \) - Refractive index \( \mu = 1.38 \) For minimum thickness, we can take \( n = 1 \): \[ t = \frac{(2 \cdot 1 - 1) \cdot 550 \times 10^{-9}}{4 \cdot 1.38} \] \[ t = \frac{1 \cdot 550 \times 10^{-9}}{4 \cdot 1.38} \] \[ t = \frac{550 \times 10^{-9}}{5.52} \] \[ t \approx 99.64 \times 10^{-9} \, \text{m} = 0.09964 \, \text{μm} \] ### Step 6: Final answer Thus, the thickness of the MgF₂ coating should be approximately: \[ t \approx 100 \, \text{nm} \]

To solve the problem of determining the thickness of the MgF₂ coating needed to produce minimum reflection for a wavelength of 550 nm, we can follow these steps: ### Step 1: Understand the condition for minimum reflection For a thin film to produce minimum reflection, the path difference between the light waves reflected from the top and bottom surfaces of the film must be an odd multiple of half the wavelength. This condition can be expressed as: \[ \Delta x = (2n - 1) \frac{\lambda}{2} \] where \( n \) is an integer (1, 2, 3, ...) and \( \lambda \) is the wavelength of light in vacuum. ...
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A glass lens is coated on one side with a thin film of magnesium fluoride (MgF_(2)) to reduce reflection from the lens surface (Fig. 2.26). The Index of refraction of MgF_(2) is 1.38, that of the glass is 1.50. What is the least coating thickness that eliminates (via interference) the reflections at the middle of the visible specturm (lambda = 550nm) ? Assume that the light is approxmately perpendicular to the lens surface.

In solar cells, a silicon solar cell (mu = 3.5) is coated with a thin film of silicon monoxide SiO (mu = 1.45) to minimize reflective losses from the surface. Determine the minimum thickness of SiO that produces the least reflection at a wavelength of 550nm, near the centre of the visible spectrum. Assume approximately normal incidence .

In solar cells, a silicon solar cell (mu = 3.5) is coated with a thin film of silicon monoxide SiO (mu = 1.45) to minimize reflective losses from the surface. Determine the minimum thickness of SiO that produces the least reflection at a wavelength of 550nm, near the centre of the visible spectrum. Assume approximately normal incidence .

A possible means for making an airplane invisible to radar is to coat the plane with an anti reflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is mu=1.5 . How thick is the oil film? Refractive index of the material of airplane wings is greater than the refractive index of polymer.

A parallel beam of light of wavelength 560 nm falls on a thin film of oil (refractive index = 1.4). What should be the minimum thickness of the film so that it strongly transmit the light ?

A glass plate (n=1.53) that is 0.485 mu m thick and surrounded by air is illuminated by a beam of white light normal to the plate. (a) What wavelengths (in air) within the limits of the visible spectrum (lambda = 400 "to" 700 nm) are intensified in the reflected beam ? (b) What wavelengths within the visible spectrum are intensified in the transmitted light?

A soap film of thickness 0.0011 mm appears dark when seen by the reflected light of wavelength 580 nm.. What is the index of refraction of the soap solution, if it is known to be between 1-2 and 1.5 ?

On face of a glass (mu = 1.50) lens is coated with a thin film of magnesium fluoride MgF_(2)(mu = 1.38) to reduce reflection from the lens surface. Assuming the incident light to be perpendicular to the lens surface. The least coating thickness that eliminates the reflection at the centre of the visible spectrum (lamda = 550 nm) is about

A thin film of a specific meterial can be used to decrease the intensity of reflected light. There is destrucive inteference of wave reflected from upper and lower surface of the film. These films are called non-reflecting or anti-reflecting coatings. The process of coating the lens or surface with non-reflecting film is called blooming as shown in figure The refracting index of coating (n_(1)) is less than that of the glass (n_(2)) . 5. magnesium fluoride (MgF_(2)) is generally use as anti-reflection coating. If refractive index of MgF_(2) is 1.25, then minimum thickness of film required is (Take lambda = 500 nm )

To ensure almost 100% transmittivity, photographic lenses are often coated with a thin layer of dielectric material, like MgF_(2)(mu=1.38) . The minimum thickness of the film to be used so that at the centre of visible spectrum (lambda = 5500 Å) there is maximum transmission.

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