Related questions
Question
In the case of diffraction by a rectangular slit of width a, it can be proved that the amplitude distribution as a function of the angular position θ is given by A = A₀ sin(u)/u, where A₀ is a constant, u = (πa sin θ)/λ and the function sin(u)/u must be replaced by 1 if u=0. As seen in class, the intensity minima occur for sin θ = nλ, n=1, 2, 3, ... (a) What is the equation that determines the positions of the intensity maxima? (b) What is the angular position θ₁ of the first maximum, excluding the central maximum
( ) (a) u = nπ, n = 1, 2, 3, … (b) sen θ₁ = λ/a
( ) (a) u = nπ/2, n = 1, 3, 5, … (b) sen θ₁= λ/2a
( ) (a) tg u - u = 0 (b) sen θ₁ = 4,49341 λ/πa
( ) (a) tg u - u = 0 (b) sen θ₁= 2λ/πa
( ) (a) tg u - u = 0 (b) sen θ₁ = 3λ/πa
( ) (a) tg u - u = 0 (b) sen θ₁ = 2λ/a
( ) (a) tg u - u = 0 (b) sen θ₁ = 3λ/a
( ) (a) tg u - u = 0 (b) sen θ₁ = 4,49341 λ/a
( ) (a) cos u - u = 0 (b) sen θ₁ = 0,7391 λ/πa
( ) (a) cos u - u = 0 (b) sen θ₁ = 2λ/πa
( ) (a) cos u - u = 0 (b) sen θ₁ = 3λ/πa
( ) (a) cos u - u = 0 (b) sen θ₁ = 2λ/a
( ) (a) cos u - u = 0 (b) sen θ₁ = 3λ/a
( ) (a) cos u - u = 0 (b) sen θ₁ = 0,7391 λ/a
Expert Solution
arrow_forward
Step 1 To determine,
(a) What is the equation that determines the positions of the intensity maxima? (b) What is the angular position θ₁ of the first maximum, excluding the central maximum.
Step by stepSolved in 2 steps
Knowledge Booster
Similar questions
- The phenomenon of diffraction occurs due to the tendency of waves to spread when they encounter an opening or obstacle. Single-slit diffraction is a phenomenon observed when a wave passes through a narrow opening, such as a slit, and spreads out after passing through it. The degree of diffraction depends on the dimensions of the aperture in relation to the wavelength of the incident wave. Statement: A slit of width a is illuminated with white light. For what value of slit width will we have the first minimum for blue light, with ? = 450 ??, appearing at ? = 25th? Question options: a) 1.50 µm b) 1.98µm c) 1.05µm d) 3.60 µm e) 0.37µmarrow_forwardA plane wave with a wavelength of 605 nm is incident normally on a single slit with a width of 3.98 x 10m. Consider waves that reach a point on a far-away screen such that rays from the slit make an angle of 2.00° with the normal. The difference in phase for waves from the top and bottom of the slit (in rad) is i rad.arrow_forwardIn an interference experiment using a monochromatic source emitting light of wavelength 1, the fringes are produced by two long, narrow slits separated by a distance d. The fringes are formed on a screen which is situated at a distance D >> d.Write down an expression for the fringe width w. Please use "*" for products (e.g. B*A), "/" for ratios (e.g. B/A) and the usual "+" and "-" signs as appropriate. Use "lambda" (without the quotes) for 1 in the equation box. For example, use d*lambda for d2. Please use the "Display response" button to check you entered the answer you expect. w=arrow_forward
- You measure three segments of the distance between a diffraction slit an the screen on which the pattern forms: x1 = (15.8 ± 0.2) cm, x2 = (6.7 ± 0.1) cm, and x3 = (11.3 ± 0.1). What is the uncertainty of the total distance x1 + x2 + x3? Group of answer choices 0.4 cm 0.5 cm 0.2 cm 0.3 cm 0.1 cmarrow_forwardSS-1 Coherent light of wavelength 675 nm passes through a narrow slit of width 0.0143 mm. The diffraction pattern is projected onto a viewing screen 1.08 m away from the slit. The intensity of the light at the center of the diffraction pattern is 175 W/m². (a) Draw a picture of the of situation descried in this problem. (b) Find the width of the central bright spot on the screen, in centimeters (cm). (c) Find the distance between the center of the diffraction pattern and the m = 4 minimum on the screen, in cm. (d) What is the intensity at a point on the screen 13.5 cm from the central maximum?arrow_forwardThe picture shows the interference pattern obtained in a double-slit experiment with monochromatic light of wavelength 650 nm. The distance shown between intensity maxima is y = 4 mm. How far away (in mm) from the central maxima is the location where the two waves have a path difference of exactly 1950 nm?arrow_forward
- DS-3 Coherent, monochromatic light goes through a pair of slits which are spaced a distance 0.600 mm apart, and the interference pattern is projected onto a screen. You notice that there are 11 bright fringes within the central diffraction maximum. What is the width of each slit (a)?arrow_forwardA monochromatic source of light of wavelength λ, is incident on a slit of width a. Obtain an expression to represent the angle of the 4th diffraction minimum. Hint: Think about Huygens construction with the single slit divided into four equal length sections acting as sources of secondary waves. Write down a general expression for the nth minimum angle. If λ =500 nm, a = 4 μm, determine the angle of the 2nd minimum.arrow_forwardCalculate the value of the position x of the 1st order diffraction peak (m= 1) relative to the middle of the screen for light of 529 nm illuminating the grating with 200 lines per mm (200 l/mm). The screen is 2 m away from the grating (see picture below). Provide your answer in SI units. Screen 一个个个 light Grating KX→arrow_forward
arrow_back_ios
arrow_forward_ios