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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)

580 nm light shines on a double slit with d 0000125 m What is the angle of the third dark interference minimum m 3 Remember nano means 109 Unit deg class=

Respuesta :

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,

[tex]d\sin\theta=(m+\dfrac{1}{2})\lambda[/tex]

Put m = 3 and other values also.

[tex]d\sin\theta=(3+\dfrac{1}{2})\lambda\\\\d\sin\theta=\dfrac{7\lambda}{2}\\\\\sin\theta=\dfrac{7\lambda}{2d}\\\\\theta=\sin^{-1}(\dfrac{7\lambda}{2d})\\\\\theta=\sin^{-1}(\dfrac{7\times 580\times 10^{-9}}{2\times 0.000125 })\\\\\theta=0.93^\circ}[/tex]

So, the angle is 0.93°.

armygirl300112

Answer: 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.

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