Question

Question: If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm ) at $\mathbf{65}\mathbf{.}\mathbf{0}\mathbf{°}$ from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm )?

Short answer

The second-order bright spot for violet light is at angle $20.2°$.

Step-by-step solution

Given Data

Given:

Angle of third order bright spot is ${\theta }_3=65.0^{\circ }$

Wavelength of red light is ${\lambda }_1=700\hspace{0.33em}nm=700\times {10}^{-9}\hspace{0.33em}m$

Wavelength for violet light is ${\lambda }_2=400\hspace{0.33em}nm=400\times {10}^{-9}\hspace{0.33em}m$

Formula used to solve the question

A diffraction grating is an optical component that has the ability to break light into its constituent wavelengths.

The angular position of the fringe is given by the following relation.

$\mathit{s}\mathit{i}\mathit{n}{\mathit{\theta }}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }}{\mathbf{d}}$

Here,$\mathit{\lambda }$ is the wavelength of light used anddis the slit width.

Determine the angle of the second-order bright spot for violet light

The slit size is given as$d=\frac{m\lambda }{\sin {\theta }_m}$is a constant since the grating is not changing.

Therefore,

$\frac{n{\lambda }_r}{\sin {\theta }_{rn}}=\frac{m{\lambda }_v}{\sin {\theta }_{vm}}$

Here, subscript states that the value is of violet light and subscript states that the value is of violet light.

Now, take, n = 3 and m = 2

$$\begin{gathered} \sin {\theta }_{v2}=\sin {\theta }_{r3}\frac{2{\lambda }_v}{3{\lambda }_r} \\ =\sin 65°\times \left(\frac{2\times 400nm}{3\times 700nm}\right) \\ =0.345 \\ {\theta }_{v2}={\sin }^{-1}\left(0.345\right) \\ =20.2° \end{gathered}$$

Thus, the second-order bright spot for violet light is at angle $20.2°$.