In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 150 cm. What is the phase difference (in degrees) between the waves from the two slits arriving at a point P when the angular distance of P is 10° relative to the central peak, and the wavelength is 500 nm? (Convert your result so the angle is between 0 and 360°.)
A) 145°
B) 155°
C) 165°
D) 135°
E) 95°

Question

Quizplus · Accepted Answer

The phase difference between the waves can be calculated using the formula $\Delta \phi = \frac{2\pi d \sin(\theta)}{\lambda}$, where $d$ is the distance between the slits, $\theta$ is the angular distance of point P, and $\lambda$ is the wavelength of the light. Here, $d = 0.2 \, \text{mm} = 2 \times 10^{-4} \, \text{m}$, $\theta = 10^\circ$, and $\lambda = 500 \, \text{nm} = 5 \times 10^{-7} \, \text{m}$. First, convert the angle to radians: $\theta = 10^\circ \times \frac{\pi}{180} = \frac{\pi}{18}$ radians. Then, plug the values into the formula:$\Delta \phi = \frac{2\pi \times 2 \times 10^{-4} \times \sin(\frac{\pi}{18})}{5 \times 10^{-7}}$This calculation gives the phase difference in radians. To convert to degrees, multiply by $\frac{180}{\pi}$. The exact calculation yields a phase difference close to 165 degrees, fitting the options provided.

In a Double Slit Experiment, the Distance Between the Slits