Q50P - University Physics with Modern Physics

Question

Your boss asks you to design a diffraction grating that will disperse the first-order visible spectrum through an angular range of 27.0°. (See Example 36.4 in Section 36.5.) (a) What must be the number of slits per centimeter for this grating? (b) At what angles will the first-order visible spectrum begin and end?

Step-by-Step Solution

Answer

(a) The number of slits per centimeter for this grating is 10172

(b) The angle of the first-order visible spectrum begins and end is $22.7° and 49.7°$

Step 1: Constructive interference for a double slit

For a double slit to produce constructive interference, the path length difference must be an integral multiple of the wavelength;

d sinθ = mλ or m = 0, 1, -1, 2, -2, 3, -3, 4, -4, …. (constructive)

here;

$$\begin{gathered} {\lambda }_v=380nm \\ {\lambda }_r=750 nm \end{gathered}$$

Step 2: Calculation

For m=1, the visible spectrum will be dispersed continuously from the first order;

$$\begin{gathered} {\theta }_v=\arcsin \left(\frac{{\lambda }_v}{d}\right) \\ to \\ {\theta }_r=\arcsin \left(\frac{{\lambda }_r}{d}\right) \end{gathered}$$

For rainbow; as  ${\lambda }_r>{\lambda }_v and {\theta }_r>{\theta }_v$ so the diffraction grating will be;

$∆\theta ={\theta }_r-{\theta }_v=27°$

Now using identity from trigonometry;

$$\begin{gathered} \sin \left({\theta }_r-{\theta }_v\right)=\sin {\theta }_r\cos {\theta }_v-\sin {\theta }_v\cos {\theta }_r \\ as \cos {\theta }_{r/v}>0 \\ \sin {\theta }_{r/v}=\frac{{\lambda }_{r/v}}{d} \end{gathered}$$

Therefore, the equation for d;

$$\begin{gathered} \sin \Delta \theta =\frac{{\lambda }_r}{d}\sqrt{1-\frac{{\lambda }_v^2}{d^2}-\frac{{\lambda }_v}{d}}\sqrt{1-\frac{{\lambda }_r^2}{d^2}}, \\ {d}^2\sin \Delta \theta ={\lambda }_r\sqrt{d^2-{\lambda }_v^2}-{\lambda }_v\sqrt{d^2-{\lambda }_r^2} \\ d=\sqrt{\frac{1}{\sin \Delta \theta }\left({\lambda }_r\sqrt{d^2-{\lambda }_v^2}-{\lambda }_v\sqrt{d^2-{\lambda }_r^2}\right)} \end{gathered}$$

The function f(x) is;

$f\left(x\right)=\sqrt{\left(\sin {\left({27}^{\circ }\right)}^{-1}\left(750\sqrt{x^2-{380}^2}-380\sqrt{x^2-{750}^2}\right)\right.}$

If $x_0$ is the solution in nanometres;

$f\left(x_0\right)=x_0!$

So $x_0$ is also known as fixed point of f.

 

Approximating the solution;

d is in the order $\mu m$ and f is defined by x>750

guessing d=1 $\mu m$ and $x_1=1000$

$$\begin{gathered} x_2=f\left(x_1\right) \\ {x}_n=f\left(x_{n-1}\right) \end{gathered}$$

As f is continuous;

$f\left(x\right)=\sqrt{\left(sin ({27}^{\circ }{)}^{-1}\left(750\sqrt{An{s}^2-{380}^2}-380\sqrt{\left.An{s}^2-{750}^2\right)}\right.\right.}$

Evaluating f at $x_1$ and then at $x_2$

$$\begin{gathered} \begin{array}{r}{x}_1=1000\\ x_2=987.145\\ x_3=983.808\\ x_4=982.959\\ x_5=982.744\\ x_6=982.690\end{array} \\ and its limit is \\ {x}_0=982.6717 \end{gathered}$$

Step 3: The number of slits per centimeter for this grating

(a) Starting with a different number and obtaining the same limit, it turns out that our initial approximation was very close to the solution, so the obtained is the separation between the slid d such that the first order visible spectrum covers the range of 27degree;

d=983nm

so, the number of slits per centimeter is

$N=\frac{1cm}{d}=10172$

Hence, the number of slits per centimeter for this grating is 10172.

Step 4: Angles the first-order visible spectrum begin and end

(b) The beginning and the end of the 1^st order spectrum are;

$\begin{array}{r}{\theta }_v=\arcsin \left(\frac{380}{983}\right)={22.7}^{\circ }\\ {\theta }_r=\arcsin \left(\frac{750}{983}\right)={49.7}^{\circ }\\ \mathrm{\Delta }\theta ={27}^{\circ }\end{array}$

Hence, the angle of the first-order visible spectrum begins and end is $22.7° and 49.7°$