In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 609 nm. an interference pattern is observed on the screen. the distance between the screen and the double-slit is 1.69 m and the distance between the two slits is 0.118 mm. a light wave propogates from each slit to the screen. what is the path length difference between the distance traveled by the waves for the sixth-order maximum (bright fringe) on the screen?

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

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