580 nm light shines on a double slit with d = 0.000125 m. what is the angle of the second bright…

Explanation:

Step1: Recall the double - slit formula

The formula for the position of bright fringes in a double - slit experiment is $d\sin\theta = m\lambda$, where $d$ is the slit separation, $\theta$ is the angle of the fringe, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the light.

Step2: Rearrange the formula for $\sin\theta$

We can rewrite the formula as $\sin\theta=\frac{m\lambda}{d}$.

Step3: Convert the wavelength to SI units

Given $\lambda = 580\ nm=580\times10^{-9}\ m$, $d = 0.000125\ m = 1.25\times 10^{-4}\ m$, and $m = 2$.

Step4: Calculate $\sin\theta$

Substitute the values into the formula: $\sin\theta=\frac{2\times580\times 10^{-9}}{1.25\times 10^{-4}}$. [ \begin{align*} \sin\theta&=\frac{1160\times 10^{-9}}{1.25\times 10^{-4}}\ &=\frac{1160}{1.25}\times10^{-9 + 4}\ & = 928\times10^{-5}\ &=0.0928 \end{align*} ]

Step5: Calculate the angle $\theta$

Since $\sin\theta = 0.0928$, then $\theta=\sin^{-1}(0.0928)$. $\theta=\sin^{-1}(0.0928)\approx5.32^{\circ}$

Answer:

$5.32$