Question

For what slit-width-to-wavelength ratio does the first minimum of a single-slit diffraction pattern appear at (a) $30°$, (b) $60°$, and (c) $90°$$?$

Short answer

a. Slit width to wavelength for$30°$is$2$.

b. Slit width to wavelength for${60}^{°}$is$1.15$.

c. Slit width to wavelength for${90}^{°}$is$1$.

Step-by-step solution

Formula for first order grating

The angles at which dark fringes is,

$\mathrm{asin}\left({\mathrm{\theta }}_{\mathrm{p}}\right)=\mathrm{p\lambda }$

Slit width to wavelength ratio is,

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{\sin \left({\mathrm{\theta }}_1\right)}$

Calculation foraλ (part a)

(a).

For $\mathrm{\theta }={30}^{°}$:

The first order grating is,

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{\sin \left({\mathrm{\theta }}_1\right)}$

Substitutelocalid="1649242325679" $\mathrm{\theta }$value,

we get,

$\sin \left({30}^{°}\right)=\frac{1}{2}$

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{(1/2)}$

$\frac{\mathrm{a}}{\mathrm{\lambda }}=2$

Calculation foraλ(part b)

(b).

For $\mathrm{\theta }={60}^{°}$:

The first order grating is,

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{\sin \left({\mathrm{\theta }}_1\right)}$

Substitutelocalid="1649242354752" $\mathrm{\theta }$value,

We get,

$\sin \left({60}^{°}\right)=\frac{\sqrt{3}}{2}$

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{(\sqrt{3}/2)}$

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{2}{\sqrt{3}}$

$\approx 1.15$

Calculation foraλ(part c)

(c).

For $\mathrm{\theta }={90}^{°}$:

The first order grating is,

$\frac{\mathrm{a}}{\mathrm{\lambda }}=\frac{1}{\sin \left({\mathrm{\theta }}_1\right)}$

$\sin \left({90}^{°}\right)=1$

$\frac{\mathrm{a}}{\mathrm{\lambda }}=1$