Question

A diffraction grating illuminated by monochromatic light normal to the grating produces a certain line at angle . (a) What is the product of that line’s half-width and the grating’s resolving power? (b) Evaluate that product for the first order of a grating of slit separation 900 nm in light of wavelength 600 nm.

Short answer

1. The product of half-width and the resolving power is $\tan \theta$.
2. The value of the product for the first order is $0.89$.

Step-by-step solution

Concept

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

$\mathbf{dsin\theta }\mathbf{=}\mathbf{m}\mathrm{\lambda }··········································\left(\mathbf{a}\right)$

The line’s half-width is given by

${\mathrm{\Delta \theta }}_{\mathrm{hw}}=\frac{\mathrm{\lambda }}{\mathrm{Ndcos\theta }}··································\left(1\right)$

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

Given Data

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

wavelength used$\lambda =600\text{nm}$

The product of half-width and the resolving power

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:

$$\begin{gathered} \Delta {\theta }_{hw}\times R=\left(\frac{\lambda }{Nd\cos \theta }\right)\left(Nm\right) \\ =\frac{m\lambda }{d\cos \theta } \\ =\frac{d\sin \theta }{d\cos \theta } \\ =\tan \theta \end{gathered}$$

Thus, the product of half-width and the resolving power is $\tan \theta$.

The value of product for first order

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

$$\begin{gathered} \Delta {\theta }_{hw}\times R=\tan \theta \\ =\frac{\sin \theta }{\cos \theta } \\ =\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }} \\ =\frac{1}{\sqrt{{\left(\frac{1}{\sin \theta }\right)}^2-1}}············································\left(3\right) \end{gathered}$$

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

$\frac{1}{\sin \theta }=\frac{d}{\lambda }$ .

Substituting the above value in equation (3),

$$\begin{gathered} \Delta {\theta }_{hw}\times R=\frac{1}{\sqrt{{\left(\frac{d}{\lambda }\right)}^2-1}} \\ =\frac{1}{\sqrt{{\left(\frac{900\text{nm}}{600\text{nm}}\right)}^2-1}} \\ =0.89 \end{gathered}$$

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