Light of wavelength 560 nm illuminates a single slit placed 60 cm from a screen.The separation between the first and third minima in the diffraction pattern is 2.8 mm.What is the slit width?

Question

Quizplus · Accepted Answer

We can use the formula for the position of minima in a single slit diffraction pattern:
$$d\sin	heta = m\lambda$$
where $d$ is the slit width, $	heta$ is the angle between the center of the slit and the minima, $m$ is the order of the minima, and $\lambda$ is the wavelength of light.

The distance between consecutive minima is given by:
$$y_{m+1}-y_m = d\left(\frac{\lambda}{D}ight)$$
where $y_m$ is the distance of the $m$th minimum from the center of the pattern, and $D$ is the distance between the slit and the screen.

We are given that the separation between the first and third minima is 2.8 mm. Therefore:
$$y_2-y_1 = d\left(\frac{\lambda}{D}ight) = 2.8	ext{ mm}$$
We can also find the angle $	heta$ corresponding to the first minimum:
$$\sin	heta = \frac{y_1}{D} = \frac{1}{2}\left(\frac{560	ext{ nm}}{60	ext{ cm}}ight) = 4.67	imes10^{-4}$$
Taking the approximation $\sin	heta \approx 	an	heta \approx 	heta$ (for small angles), we get:
$$	heta = 4.67	imes10^{-4}	ext{ rad}$$
Using these values in the formula for $d$, we find:
$$d = \frac{y_2-y_1}{\lambda	an	heta} = \frac{2.8	ext{ mm}}{560	ext{ nm}	imes 4.67	imes10^{-4}} \approx 0.24	ext{ mm}$$
Therefore, the answer is (B).

Light of Wavelength 560 Nm Illuminates a Single Slit Placed