91影视

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

d sin胃 = m位 or m = 0, 1, -1, 2, -2, 3, -3, 4, -4, 鈥�. (constructive)

here;

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

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

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

For rainbow; asrole="math" localid="1668221070342" ${位}_r>{位}_{v}and{胃}_r>{胃}_v$ so the diffraction grating will be;

$鈭�胃={胃}_r-{胃}_v=27掳$

Now using identity from trigonometry;

$$\begin{gathered} \sin \left({胃}_r鈭�{胃}_v\right)=\sin {胃}_r\cos {胃}_v鈭�\sin {胃}_v\cos {胃}_r \\ as\cos 鈦�{胃}_{r/v}>0 \\ \sin {胃}_{r/v}=\frac{{位}_{r/v}}{d} \end{gathered}$$

Therefore, the equation for d;

$$\begin{gathered} \sin 螖胃=\frac{{位}_r}{d}\sqrt{1鈭�\frac{{位}_v^2}{d^2}鈭�\frac{{位}_v}{d}}\sqrt{1鈭�\frac{{位}_r^2}{d^2}}, \\ {d}^2\sin 螖胃={位}_r\sqrt{d^2鈭�{位}_v^2}鈭�{位}_v\sqrt{d^2鈭�{位}_r^2} \\ d=\sqrt{\frac{1}{\sin 螖胃}\left({位}_r\sqrt{d^2鈭�{位}_v^2}鈭�{位}_v\sqrt{d^2鈭�{位}_r^2}\right)} \end{gathered}$$

The function f(x) is;

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

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 $渭m$and f is defined by x>750

guessing d=1$渭m$ 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;

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

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

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

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.

$\begin{array}{r}{胃}_v=\arcsin \left(\frac{380}{983}\right)={22.7}^{鈭�}\\ {胃}_r=\arcsin \left(\frac{750}{983}\right)={49.7}^{鈭�}\\ \mathrm{螖}胃={27}^{鈭�}\end{array}$

Hence, the angle of the first-order visible spectrum begins and end is $22.7掳and49.7掳$