Question

(a) How wide is a single slit that produces its first minimum for $633-\mathrm{n}\mathrm{m}$light at an angle of $28{.0}^{\circ }$? (b) At what angle will the second minimum be?

Short answer

(a) The slit is a certain width $1.35\mu \mathrm{m}$.

(b) $69{.7}^{\circ }$ is the angle at which the second minimum occurs.

Step-by-step solution

Definition of wavelength

The wavelength of a waveform signal conveyed in space or down a wire is the distance between identical points (adjacent crests) in adjacent cycles.

Given Data

Angle for the first minimum is$28{.0}^{\circ }$

The wavelength of the light is,

$\lambda =633nm\left(\frac{10^{-9}\mathrm{m}}{1nm}\right)=6.33\times 10^{-7}\mathrm{m}$

Find the width of the slit

To find the width of the slit, solve the equation$\mathrm{D}\mathrm{s}\mathrm{i}\mathrm{n}(\theta )=\mathrm{m}\lambda$for D, where$\mathrm{m}=1$.

$\begin{array}{c}\mathrm{D}=\frac{\mathrm{m}\lambda }{\mathrm{s}\mathrm{i}\mathrm{n}(\theta )}\\ =\frac{1\times \left(6.33\times 10^{-7}\mathrm{m}\right)}{\mathrm{s}\mathrm{i}\mathrm{n}\left(28{.0}^{\circ }\right)}\\ =1.35\times 10^{-6}\mathrm{m}\left(\frac{1\mu m}{10^{-6}\mathrm{m}}\right)\\ =1.35\mu \mathrm{m}\end{array}$

Therefore, the slit is a certain width $1.35\mu \mathrm{m}$.

At what angle will the second minimum be

To calculate the angle for the first minimum, we solve the$\mathrm{D}\mathrm{s}\mathrm{i}\mathrm{n}(\theta )=\mathrm{m}\lambda$for,$\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{e}\mathrm{i}\mathrm{t}\mathrm{f}\mathrm{o}\mathrm{r}\theta \mathrm{w}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{m}=2$.

$\begin{array}{c}\theta ={\sin }^{-1}\left(\frac{2\times 633\times 10^{-9}\mathrm{m}}{1.35\times 10^{-6}\mathrm{m}}\right)\\ =69{.7}^{\circ }\end{array}$

Therefore,$69{.7}^{\circ }$is the angle at which the second minimum occurs.