Answered: In a Young's double slit experiment set… | bartleby

Related questions

Question

In a Young's double slit experiment set up, sources of wavelength 500nm illuminates two slits S,
and S, which act as two coherent sources. The Source S oscillates about its own position
according to the equation y = 0.5 sin where y is in mm and t in seconds. The minimum value
of time t for which the intensity at point P on the screen exactly infront of the upper slit
becomes minimum is
1mm
4-1m-→
2m

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Blurred answer

Similar questions

In a single slit diffraction, what is the angular width of the central bright region? Where, the wavelength of incident light is 550 nm and the width of slit w= 0.2 mm. O 0.45° O 0.17° 0.086° O 0.315° O 0.13º
= 6. Go mmh Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is AA = 645 nm. The other wavelength is AB and is unknown. On a viewing screen, the light with wavelength ^A 645 nm produces its third-order bright fringe at the same place where the light with wavelength AB produces its fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown wavelength?
A student performs a double-slit interference experiment. The student records the fact that a third-order dark spot is located at an angle of 20.11 for a given light source. The wavelength of light is then decreased by 40.0nm, and the student records the fact that the second-order dark spot is now located at an angle of 11.17. What is the slit separation?
The key aspect of two-slit interference is the dependence of the total intensity at point O on the angle θ. Find this intensity I(θ). The formula for intensity is I=(ϵ0c(amplitudeofE)2)/2 Express your answer in terms of Imax, θ, d, and λ, where Imax=2ϵ0cE(r)2. Note: cos2x should be coded as cos(x)^2. In order to make the math as simple as possible, we will define two phases: ϕ=(2π/λ)*(r−ct) and δϕ=(π/λ)dsin(θ) Then Φlower=ϕ+δϕ and Φupper=ϕ−δϕ E=2E(r)cosϕcosδϕ
Light of wavelength λ = 610 nm and intensity I0 = 240 W/m2 passes through a slit of width w = 4.8 μm before hitting a screen L = 1.6 meters away. Part (a) Use the small-angle approximation to write an equation for the phase difference, β, between rays that pass through the very top and very bottom of the slit when the rays hit a point y = 46 mm above the central maximum. Part (b) Calculate this phase difference, in radians? Part (c) What is the intensity of the light, in watts per square meter, at this point?
Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 1.30 µC charge. (Let A = 1.30 µC, B = 6.80 µC, and C = -4.32 µC.) %3D В 0.500 m 60.0° + A C magnitude N direction ° below the +x-axis
A 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.
Monochromatic, in phase light of wavelength > 752nm is incident on a piece of metal with either one or two slits on it - it is difficult to see to determine which. The diffraction pattern is observed on a viewing screen that is located 55cm away from the metal plate. The first two minima are measured to be 2.3cm and 7.0cm away from the central maxima. What is the brightness of the light at a distance of 6.3cm from the central maxima, relative to the brightness of the central maxima? x10-3 Note: Brightness is proportional to the SQUARE of the amplitude. For double slit, this means I = I (cos(d sin ) º)) ². 2 I = I₂ sin (5) xd sin 0 . For single slit, this means In both of these, I, is the maximum brightness, and d is the characteristic spacing of the slit(s).
Sound waves with frequency 2600 Hz and speed 343 m/s diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length 100 m. The opening, which has a horizontal width of 38.7 cm, faces a wall 100 m away (the figure below). Along that wall, how far from the central axis of that wall in meters will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.)
A technician is performing Young's double-slit experiment for his supervisor. He directs a beam of single-wavelength light to a pair of parallel slits, which are separated by 0.132 mm from each other. The portion of this light that passes through the slits goes on to form an interference pattern upon a screen, which is 4.50 meters distant.The light is characterized by a wavelength of 590 nm. (a)What is the optical path-length difference (in µm) that corresponds to the fifth-order bright fringe on the screen? (This is the fifth fringe, not counting the central bright band, that one encounters moving from the center out to one side.) ?µm (b)What path-length difference (in µm) corresponds to the fifth dark fringe that one encounters when moving out to one side of the central bright fringe? ?µm
Two slits separated by a distance of d=0.150 mm are located at a distance of D=810. mm from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a coherent light source with a wavelength of λ=549×10−6 mm. The interference pattern shows a peak at the center of the screen (m=0) and then alternating minima and maxima. (a) What is the path length difference in millimeters between the two waves from the two slits at the first (m=1) maximum on the screen? (b) What is the path length difference in millimeters between the two waves from the two slits at the first (m=0) minimum on the screen? (c) Calculate the distance on the screen between the central maximum (m=0) and the first (m=1) maximum. You can assume sinθ≈tanθ≈θ, with θ expressed in radians. Give your answer in millimeters.
Suppose light with wavelength 540 nm has parallel wavefronts incident on two slits that are both 0.36 mm wide, and separated by 0.5 mm. A screen is placed 1.2 m from the slit. The intensity at the optical axis is 5 mW. What is the distance from the optical axis to the second minimum on the screen in mm?

SEE MORE QUESTIONS