Question

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the centerline does the second bright fringe occur?

Answer 1

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light  

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$, hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$:

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$   (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$   (5)

Finally:

$\theta_{2}=1.8\°$   (6)