91Ó°ÊÓ

In the diffraction grating experiment, a maxima is achieved at the points where the path difference is an integral multiple of the wavelength.

$\mathbf{»å²õ¾\pm ²Ôθ}\mathbf{=}\mathbf{m}\mathrm{λ}··········································\left(\mathbf{a}\right)$

The line’s half-width is given by

${\mathrm{Δθ}}_{\mathrm{hw}}=\frac{\mathrm{λ}}{\mathrm{\pm ·»å³¦´Ç²õθ}}··································\left(1\right)$

Here, $\mathrm{N}$ is the number of rulings on the grating.

Slit separation$d=900\text{nm}$

wavelength used$\mathrm{λ}=600\text{nm}$

It is known that the resolving power $\left(R\right)$is given as-

$R=Nm··············································\left(2\right)$

Using equations .$\left(1\right),\left(2\right)\text{and}\left(\text{a}\right)$ The product of the half-width and the resolving power is given by:

$$\begin{gathered} \mathrm{Δ}{\mathrm{θ}}_{hw}×R=\left(\frac{\mathrm{λ}}{Nd\cos \mathrm{θ}}\right)\left(Nm\right) \\ =\frac{m\mathrm{λ}}{d\cos \mathrm{θ}} \\ =\frac{d\sin \mathrm{θ}}{d\cos \mathrm{θ}} \\ =\tan \mathrm{θ} \end{gathered}$$

Thus, the product of half-width and the resolving power is $\tan \mathrm{θ}$.

The value of the product for the first order can be obtained as follows:

$$\begin{gathered} \mathrm{Δ}{\mathrm{θ}}_{hw}×R=\tan \mathrm{θ} \\ =\frac{\sin \mathrm{θ}}{\cos \mathrm{θ}} \\ =\frac{\sin \mathrm{θ}}{\sqrt{1-{\sin }^2\mathrm{θ}}} \\ =\frac{1}{\sqrt{{\left(\frac{1}{\sin \mathrm{θ}}\right)}^2-1}}············································\left(3\right) \end{gathered}$$

For the first order, using equation (a) we get,

$\frac{1}{\sin \mathrm{θ}}=\frac{d}{\mathrm{λ}}$ .

Substituting the above value in equation (3),

$$\begin{gathered} \mathrm{Δ}{\mathrm{θ}}_{hw}×R=\frac{1}{\sqrt{{\left(\frac{d}{\mathrm{λ}}\right)}^2-1}} \\ =\frac{1}{\sqrt{{\left(\frac{900\text{nm}}{600\text{nm}}\right)}^2-1}} \\ =0.89 \end{gathered}$$

Thus, the value of product for the first order is $0.89$.