light of wavelength 619 nm passes through a double slit, and makes its first maximum (m = 1) at an angle of…

Explanation:

Step1: Recall double - slit formula

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

Step2: Rearrange the formula for $d$

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

Step3: Convert units

The wavelength $\lambda = 619\ nm=619\times10^{-9}\ m$. The angle $\theta = 0.226^{\circ}$. First, convert the angle to radians: $\theta_{rad}=\theta\times\frac{\pi}{180}=0.226\times\frac{\pi}{180}\ rad$. Then $\sin\theta=\sin(0.226^{\circ})\approx0.226\times\frac{\pi}{180}$ (since for small angles $\sin\theta\approx\theta$ in radians). $m = 1$.

Step4: Calculate $d$ in meters

Substitute the values into the formula: $d=\frac{1\times619\times 10^{-9}}{\sin(0.226^{\circ})}$. $\sin(0.226^{\circ})\approx0.226\times\frac{\pi}{180}\approx3.94\times10^{-3}$. So $d=\frac{619\times10^{-9}}{3.94\times10^{-3}}\ m$.

Step5: Convert $d$ to millimeters

$d=\frac{619\times10^{-9}}{3.94\times10^{-3}}\times10^{3}\ mm$. $d=\frac{619\times10^{-9 + 3}}{3.94\times10^{-3+3}}\ mm=\frac{619\times10^{-6}}{3.94}\ mm\approx0.157\ mm$.

Answer:

$0.157$