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)?

Answer 1

Answer:

0.796 degree

Explanation:

We are given that

Wavelength=$\lambda=580 nm=580\times 10^{-9} m$

$1nm=10^{-9} m$

$d=0.000125 m$

We have .to find the angle of the third dark interference minimum=3

$sin\theta=\frac{m\lambda}{d}$

Using the formula

$sin\theta=\frac{3\times 580\times 10^{-9}}{0.000125}$

$sin\theta=0.0139$

$\theta=sin^{-1}(0.0139)$

$\theta=0.796^{\circ}$

Hence, the angle of the third dark interference minimum=0.796 degree

Answer 2

Answer:

Explanation:

wavelength, λ= 580 nm

d = 0.000125 m

m = 3

Let the angle is θ.

d sin θ = m λ

0.000125 Sinθ = 3 x 580 x 10^-9

Sinθ = 4.64 x 10^-3

θ = 0.266°