If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.0 m away

0.82 mm

Explanation:

The formula for calculation an bright fringe from the central maxima is given as:

so for the distance of the second-order fringe when wavelength = 745-nm can be calculated as:

where;

n = 2

= 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

= 0.00276 m

= 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength = 660-nm is as follows:

= 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

= 10 ⁻³ (2.76 - 1.94)

= 10 ⁻³ (0.82)

= 0.82 × 10 ⁻³ m

=  0.82 × 10 ⁻³ m

= 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

the components of a vector are the parts of the vectors that are to each other.

b perpendicular