Question

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)

Answer 1

Answer: 0.665 deg

Step-by-step 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.

Answer 2

Step-by-step 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,

`d\sin\theta=(m+(1)/(2))\lambda`

Put m = 3 and other values also.

`d\sin\theta=(3+(1)/(2))\lambda\\\\d\sin\theta=(7\lambda)/(2)\\\\\sin\theta=(7\lambda)/(2d)\\\\\theta=\sin^(-1)((7\lambda)/(2d))\\\\\theta=\sin^(-1)((7* 580* 10^(-9))/(2* 0.000125 ))\\\\\theta=0.93^\circ}`

So, the angle is 0.93°.