An EM wave with wavelength 6.50E-7 m passes through a thin slit and form a diffraction pattern on a screen 3.67 m away. The distance between the central bright fringe and the first dark fringe is measured to be 4.20-3 m. What is the slit width (in meter)?

What light of wavelength 440 nm passes through a single slit, the width of the central maximum (the angles between the two zeros on each side of it) is found to be 73.74 degrees. What is the width of the slit?

Light with a wavelength of 633 nm goes through a narrow slit. The angle between the first minimum on one side of the central maximum and the first minimum on the other side is 1.2o. What is the width of the slit?

In a single-slit diffraction pattern, the central fringe is 320 times as wide as the slit. The screen is 17,000 times farther from the slit than the slit is wide. What is the ratio /W, where is the wavelength of the light shining through the slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that sin is approximately equal to tan .

A flat screen is located 0.45 m away from a single slit. Light with a wavelength of 540 nm (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is 0.046 m. What is the width of the slit?

620-nm light is shined on two narrow slits separated by 0.2-mm. What is the distance between two adjacent bright fringes on a screen at a distance 5.5-m from the slits? The distance, Ay = Unit cm At what angle from the centerline does the 2nd dark fringe occur? The angle, 0₂ = Unit degree

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Two radio antennas are separated by 1.60 m. Both broadcast identical 750 MHz waves. If you walk around the antennas in a circle of radius 10.0 m, how many maxima will you detect?

A slit of width 0.15 mm is located 10.0 m from a screen. Light with a wavelength of 450 nm passes through this slit. Determine the distance between the first and the third dark fringes on the screen.

There is a thin film of soap bubble of thickness d on air with a refractive index of n. When an incident light of wavelength is falling on the soap bubble it is creating a dark fringe and when the wavelength is falling it is creating a bright fringe of the same order at the same position on screen. Prove that the order of the fringe is From the above relation find the thickness of the soap bubble of refractive index 1.35 which is creating a bright fringe for 370nm and a dark fringe for 500nm.