Intensity of double-slit pattern: In a double slit experiment, the intensity of light at the center of the central bright fringe is measured to be 6.2 µW/m². What is the intensity halfway between the center of this fringe and the first dark band, assuming that the small-angle approximation is valid? A) 6.2 µW/m² B) 3.1 µW/m² C) 4.7 µW/m² D) 1.6 µW/m² E) 0.12 µW/m²

Question

Quizplus · Accepted Answer

The intensity of the double-slit pattern is given by:
$I = I_0\frac{\sin^2\left(\frac{\pi d \sin	heta}{\lambda}ight)}{\left(\frac{\pi d \sin	heta}{\lambda}ight)^2}$,
where $I_0$ is the intensity of the light source, $d$ is the distance between the slits, $\lambda$ is the wavelength of the light, and $	heta$ is the angle between the center of the central bright fringe and the point where the intensity is being measured.

Assuming the small-angle approximation is valid, we have $\sin	heta \approx 	heta$, where $	heta$ is in radians. Let $x$ be the distance between the center of the central bright fringe and the point where the intensity is being measured. Then, half of this distance is $\frac{x}{2}$. The angle corresponding to this distance is given by $	heta = \frac{\frac{x}{2}}{L}$, where $L$ is the distance from the double-slit to the screen.

Therefore, the intensity halfway between the center of the central bright fringe and the first dark band is given by:
$I = I_0\frac{\sin^2\left(\frac{\pi d}{\lambda}\frac{x}{2L}ight)}{\left(\frac{\pi d}{\lambda}\frac{x}{2L}ight)^2}$.

Since the intensity at the center of the central bright fringe is given to be $6.2 µW/m^2$, we have:
$I_0 = 6.2 µW/m^2$,
$x = \frac{d\lambda L}{2d} = \frac{\lambda L}{2}$.

Substituting these values, we get:
$I = 3.1 µW/m^2$.
Therefore, the answer is B.

Intensity of Double-Slit Pattern: in a Double Slit Experiment, the Intensity