Question

Monochromatic light (wavelength$\mathbf{=}\mathbf{450}\mathbf{}\mathit{n}\mathit{m}$) is incident perpendicularly on a single slit (width$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}\mathit{m}$). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is $\mathbf{1}\mathbf{.}\mathbf{8}\mathit{m}\mathit{m}$.

(a) What is the distance from the slit to the screen? (Hint:The angle to either minimum is small enough that$\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{\approx }\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }$.)

(b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?

Short answer

(a) The distance of the screen from the slit is$0.8m$ .

(b) The distance between the first and third minima is $1.8mm$.

Step-by-step solution

Given data

Slit width $a=0.4mm$

Distance between the two minima on either side of the central maximalocalid="1663048702619" $∆y_{1,-1}=1.8mm$

Wavelength of incident light $\lambda =450nm$

Definition and concept of diffraction from a grating

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance $\theta$of the $m_{th}$ order minima in diffraction pattern produced from a single slit having slit width $a$ is

$\mathit{a}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …(i)

Here, $\lambda$ is the wavelength of the incident light.

(a) Determining the distance of the screen from the slit

Let the distance from the slit to the screen be$D$ . For small angular distances

$$\begin{gathered} \sin \theta \approx \tan \theta \\ =\frac{y}{D} \end{gathered}$$

Here $y$ is the distance measured on the screen. Thus, from equation (i) the separation between the first two minima $m\pm 1$ on either sides of the central maxima is

$$\begin{gathered} \frac{a∆y_{1,-1}}{D}=\left\{1-\left(-1\right)\right\}\lambda \\ D=\frac{a∆y_{1,-1}}{2\lambda } \end{gathered}$$

Substitute the values to get

$$\begin{gathered} D=\frac{0.4mm\times 1.8mm}{2\times 450nm} \\ =\frac{0.4\times {10}^{-3}m\times 1.8\times {10}^{-3}m}{2\times 450\times {10}^{-9}m} \\ =0.8m \end{gathered}$$

Thus, the distance is$0.8m$ .

(b) Determining the distance between the first and third minima

From equation (i), the distance between the first $\left(m=1\right)$ and third minima$\left(m=3\right)$ is

$$\begin{gathered} \frac{a∆y_{1,3}}{D}=\left\{3-1\right\}\lambda \\ ∆y_{1,3}=\frac{2\lambda D}{a} \\ =\frac{2\times 450\times {10}^{-9}m\times 0.8m}{0.4\times {10}^{-3}m} \\ =1.8mm \end{gathered}$$

Thus, the distance is $1.8mm$.