580 nm light shines on a double slit with d = 0.000125 m. What is the
angle of the third dark interference
minimum (m = 3)?
(Remember, nano means 10-9.)
(Unit = deg)

Explanation:

Given that,

The wavelength of light = 580 nm

Slit separation, d = 0.000125 m

We need to find the angle of the third dark interference. For the dark fringe,

Put m = 3 and other values also.

So, the angle is 0.93°.

0.665 deg

Explanation:

m=3

lambda= 580

d (converted to nanometers)= 125000

Using the equation of angle=arcsine of m-1/2 times lambda divided by d, filled in it would be arcsine of 3-1/2 times 580 over 125000.

mass and volume are the answers you are looking for

no

explanation:

no i dont know how to solve this you should probably try to google the answer