Monochromatic light with wavelength A = 680nm falls vertically on an optical grating with N = 100 lines per centimetre. light with wavelength A = 680nm. Calculate the diffraction angle ag for the 8th order maximum. Give the result in degrees.
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- Light from a sodium light source is incident perpendicular to a transmission grating that has 4,350 lines per centimeter. Sodium light consists of two very close wavelengths (2₁= 589.0 nm and 22 = 589.6 nm). The resulting diffraction pattern is displayed on a projection screen 1.60 m from the grating and for a particular order m, the maxima for these two wavelengths is separated by 1.32 mm. Determine the order for these two maxima. (Enter a whole number.) Submit AnswerA He-Ne gas laser which produces monochromatic light of a known wavelength ? =6.35 ? 10^−7 ? is used to calibrate a reflection grating in a spectroscope. The first-orderdiffraction line is found at an angle of 22 degrees to the incident beam. How many lines per meter are there on the grating?A telescope has a circular aperture of diameter D = 4.4 m. A light with wavelength λ = 690 nm travels through the telescope. Express the limiting angle of resolution, θmin, in terms of λ and D. You may assume that θmin is very small. θmin =
- A transmission grating with d=1.5μm is illuminated at various angles of incidence by light of wavelength 600nm. Derive an expression for the angular deviation of the first order diffracted beam (m=1) compared to the 0th order (m=0) as a function of the angle of incidence. Plot as a function of angle of incidence (0 to 90 degrees) the angular deviation of the first order diffracted beam from the 0th order (m=0) beam.When the light of a wavelength λ = 561 nm is incident on a diffraction grating the first maximum after the center one is found to occur at an angle of θ1 = 7.5 degrees. Part (a) Write an expression for the grating's line density, n, in terms of the given quantities. Part (b) Calculate this spacing in lines per centimeter? Part (c) Find the angle of the second-order intensity maximum, θ2, in degrees.A single slit of width 0.11 mm is illuminated by a mercury light of wavelength 543576 nm. Find the intensity in terms of Io. at a 12° angle to the axis in terms of the intensity of the central maximum. Round your answer to 7 decimal places.
- The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter is 2.41 mm? Assume light with a wavelength of λ = 550 nm. (Note: The distance from the pupil to the retina is 25.4 mm. In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is n = 1.336.)Hint: The size of the spot is twice the distance from the main axis to the first minimum.The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter is 2.74 mm? Assume light with a wavelength of λ = 550 nm. (Note: The distance from the pupil to the retina is 25.4 mm. In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is n = 1.336.)Hint: The size of the spot is twice the distance from the main axis to the first minimumress SA A thin film of magnesium fluoride (n = 1.38) is applied to glass (n = 1.50) with a thickness of 100 nm. For what visible wavelength of light will this coating act as nonreflecting? Assume normal incidence. O 491 nm O 600 nm O 504 nm 552 pm Q Search 15 40 %5 6 Il app.honorlock.com is sharing your screen. Stop, sharing Hide 99+ 3+ hp 103 a malip 8
- A diffraction grating has 2500 lines/cm. What is the angle between the first-order maximum for red light wavelentgh = 680 nm and the first-order maximum for blue light wavelength = 410 nm?Problem 2. Suppose that a Michelson interferometer is illuminated by a source emitting a doublet of vacuum wavelengths, 21 and 22. As one mirror is moved, the fringes periodically disappear and then reappear. If a displacement Ad of the mirror causes a one-cycle variation in fringe visibility, derive an expression for Ad in terms of Aλ = 2-1. Then calculate Ad for the sodium doublet, where λ₁ = 588.995 nm and 2 = 589.592 nm.The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter is 2.62 mm? Assume light with a wavelength of λ = 550 nm. (Note: The distance from the pupil to the retina is 25.4 mm. In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is n = 1.336.)Hint: The size of the spot is twice the distance from the main axis to the first minimum.