Q3E - University Physics with Modern Physics

Question

Question: Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin u can be? What does this tell you is the largest that m can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Step-by-Step Solution

Answer

1. The total number of dark fringes or minima on the screen would be 226.
2. The farthest dark fringe will make an angle of 83 degrees from horizontal

Step 1: Given Data

Diffraction is a process by which a beam of light spreads when passing through a narrow passage or across the edge of an obstacle accompanied by interference between the waveforms.

Step 2: Total number of dark fringes

It is given that,

Wavelength of light source, λ = 585 nm = 585 x 10^-9 m

Width of the slit, a = 0.0666 mm = 6.66 x 10^-5 m

 

 

By using the equation:

$$\begin{gathered} \sin \theta =\frac{m\lambda }{a} \\ a=\frac{m\lambda }{\sin \theta } \end{gathered}$$

 

It is evident that the largest value of angle is 90 degrees therefore,

 

$$\begin{gathered} \sin \theta =\frac{m\lambda }{a} \\ \sin {\theta }_{max}=\frac{m\lambda }{a} \\ m=\frac{a \sin {\theta }_{max}}{\lambda } \\ m=\frac{a \sin 90°}{\lambda } \\ =\frac{6.66*{10}^{-5}*1.0}{585*{10}^{-9}} \\ =113.8 \end{gathered}$$

 

Thus, the farthest dark fringe from central maxima is at m = 113. These are dark fringes on one side of central maxima and total number of dark fringes on the screen would be

 

N = 113*2

    = 226

 

Therefore, total number of dark fringes or minima on the screen would be 226.

Step 3: Angle of farthest dark fringe

The farthest dark fringe is 113rd dark fringe from the central maxima and its angle from horizontal can be found using 

 

$$\begin{gathered} \sin \theta =\frac{113 \lambda }{a} \\ \theta ={\sin }^{-1}\left(\frac{113\lambda }{a}\right) \\ ={\sin }^{-1}\left(\frac{113*585*{10}^{-9}}{6.66*{10}^{-5}}\right) \\ =83.0° \end{gathered}$$

 

Therefore, the farthest dark fringe will make an angle of 83 degrees from horizontal.