Question

quick refresher
this refresher from 35.2 youngs interference experiment will help you get familiar with prerequisite concepts
a double - slit arrangement in air produces interference fringes for sodium light ($\lambda = 589$ nm) that are $0.20^\circ$ apart. what is the angular separation if the entire arrangement is immersed in water ($n = 1.33$)?
$\bigcirc 0.35^\circ$
$\bigcirc 0.20^\circ$
$\bigcirc 0.11^\circ$
$\bigcirc 0.15^\circ$
$\bigcirc 0.27^\circ$

Explanation:

Step1: Recall Double - Slit Interference Formula

In double - slit interference, the angular separation \(\theta\) between adjacent fringes is given by \(\theta\approx\sin\theta = \frac{\lambda}{d}\) (for small angles, \(\sin\theta\approx\theta\) in radians, but we can work with degrees here as we are dealing with the ratio). When the medium changes, the wavelength of light changes. The wavelength of light in a medium with refractive index \(n\) is \(\lambda_n=\frac{\lambda}{n}\), where \(\lambda\) is the wavelength in vacuum (or air, since the refractive index of air is approximately 1).

Let \(\theta_1\) be the angular separation in air and \(\theta_2\) be the angular separation in water. We know that \(\theta_1=\frac{\lambda}{d}\) and \(\theta_2 = \frac{\lambda_n}{d}=\frac{\lambda/(n)}{d}=\frac{1}{n}\cdot\frac{\lambda}{d}\). So, \(\theta_2=\frac{\theta_1}{n}\).

Step2: Substitute the Given Values

We are given that \(\theta_1 = 0.20^{\circ}\) and \(n = 1.33\). Substituting these values into the formula for \(\theta_2\), we get \(\theta_2=\frac{0.20^{\circ}}{1.33}\approx0.15^{\circ}\)? Wait, no, wait. Wait, the formula is \(\theta\propto\lambda\), and \(\lambda\) in water is \(\lambda_w=\frac{\lambda_{air}}{n}\). So \(\theta_2=\frac{\lambda_w}{d}=\frac{\lambda_{air}/n}{d}=\frac{1}{n}\cdot\frac{\lambda_{air}}{d}=\frac{\theta_1}{n}\). Wait, \(\theta_1 = 0.20^{\circ}\), \(n = 1.33\), so \(\theta_2=\frac{0.20}{1.33}\approx0.15^{\circ}\)? Wait, no, let's recalculate: \(0.20\div1.33\approx0.1504^{\circ}\approx0.15^{\circ}\)? Wait, but let's check again.

Wait, the formula for the angular separation between fringes in double - slit interference is \(\Delta\theta=\frac{\lambda}{d}\) (for small angles, where \(\Delta\theta\) is the angular separation, \(\lambda\) is the wavelength of light, and \(d\) is the slit separation). When the arrangement is in water, the wavelength of light is \(\lambda'=\frac{\lambda}{n}\), where \(n\) is the refractive index of water. So the new angular separation \(\Delta\theta'=\frac{\lambda'}{d}=\frac{\lambda/(n)}{d}=\frac{1}{n}\cdot\frac{\lambda}{d}=\frac{\Delta\theta}{n}\).

So \(\Delta\theta'=\frac{0.20^{\circ}}{1.33}\approx0.15^{\circ}\). Wait, but let's do the division: \(0.20\div1.33 = 0.20\div\frac{4}{3}=0.20\times\frac{3}{4}=0.15\). Yes, so the angular separation in water is \(\frac{0.20^{\circ}}{1.33}\approx0.15^{\circ}\)? Wait, no, 0.20 divided by 1.33: 1.33 times 0.15 is 0.1995, which is approximately 0.20. Wait, no, 0.20 / 1.33: 1.33*0.15 = 0.1995, so 0.20/1.33≈0.15. Wait, but let's check the options. The options are \(0.35^{\circ}\), \(0.20^{\circ}\), \(0.11^{\circ}\), \(0.15^{\circ}\), \(0.27^{\circ}\). So the correct calculation is \(\theta_2=\frac{\theta_1}{n}=\frac{0.20^{\circ}}{1.33}\approx0.15^{\circ}\).

Answer:

D. \(0.15^{\circ}\) (assuming the option with \(0.15^{\circ}\) is the correct one, based on the calculation \(\theta_2=\frac{0.20^{\circ}}{1.33}\approx0.15^{\circ}\))