Q26E - University Physics with Modern Physics

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 )?

Step-by-Step Solution

Answer

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

Step 1: 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 ${\lambda }_2=400\hspace{0.33em}nm=400\times {10}^{-9}\hspace{0.33em}m$  

Step 1: 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 and  d is the slit width.

Step 2: 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

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

  

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