A screen is illuminated by 411 nm light as shown in the figure below. The distan

Question

A screen is illuminated by 411 nm light as shown in the figure below. The distance from the slits to the screen is 6.4 m. Figure: Not drawn to scale. Find the minimum positive phase angular value such that I/I0 = 46%, 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 ; i.e., an angle within the central maximum. Answer in units of degree

Explanation / Answer

009)

I/I0 = cos2(/2)

0.46 = cos2(/2)

= 94.6 degrees

010)

= /2 = 411*1.65088/(2*3.1416) = 108 nm

011)

= d sin

sin = /d = 108e-9/0.82e-3

= 0.007546 degrees

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