A. B. C. A screen is illuminated by 676 nm light as shown in the figure below. T

Question

A.

B.

C.

A screen is illuminated by 676 nm light as shown in the figure below. The distance from the slits to the screen is 4.8 m. Figure: Not drawn to scale. Find the minimum positive phase angular value such that I/I0 = 77%, where I0 is the intensity at the central maximum and I is the intensity at the position y on the screen. Answer in units of degree Find the difference in path length for the rays from the two slits. Answer in units of nm Find the minimum positive angular value theta; i.e., an angle within the central maximum. Answer in units of degree

Explanation / Answer

A) = 2cos-1((I/Io)) = 2cos-1(sqrt(0.77)) = 2cos-1(0.8775) = 1.00036 rad = 57.31640

B) = ()/2 = 1.07627 * 10-7 m = 107.627 nm

C) = sin-1(()/(2d)), d = 0.75mm

=> = sin-1(0.000144) = 0.000144 rad = 0.0082220

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