A single thin slit forms a diffraction pattern when illuminated with monochromatic light.The fourth minimum of the pattern occurs at an angle of 32.0° away from the central maximum.At what angle does the fifth minimum occur?
A)41.5°
B)41.0°
C)40.5°
D)42.0°
E)42.5°

Question

Quizplus · Accepted Answer

The angle for the minima in a single slit diffraction pattern is given by $\sin(\theta) = \frac{m\lambda}{a}$, where $m$ is the order of the minimum, $\lambda$ is the wavelength of the light, and $a$ is the width of the slit. For two different minima, $m_1$ and $m_2$, with angles $\theta_1$ and $\theta_2$, the equation can be set up for each and then divided to eliminate $\lambda/a$, giving $\frac{\sin(\theta_1)}{m_1} = \frac{\sin(\theta_2)}{m_2}$. Given the fourth minimum occurs at 32.0°, we can find the angle for the fifth minimum by setting $m_1 = 4$, $\theta_1 = 32.0°$, and $m_2 = 5$, and solving for $\theta_2$. This approach leads to the calculation of the angle for the fifth minimum, which is approximately 41.5°, option A.

A Single Thin Slit Forms a Diffraction Pattern When Illuminated