Q40P - Fundamentals Of Physics

Question

Figure 36-45 gives the parameter $\mathit{\beta }$ of Eq. 36-20 versus the sine of the angle in a two-slit interference experiment using light of wavelength 435 nm. The vertical axis scale is set by ${\mathit{\beta }}_{\mathbf{s}}\mathbf{=}\mathbf{80}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{rad}}$. What are (a) the slit separation, (b) the total number of interference maxima (count them on both sides of the pattern’s center), (c) the smallest angle for a maxima, and (d) the greatest angle for a minimum? Assume that none of the interference maxima are completely eliminated by a diffraction minimum.

Step-by-Step Solution

Answer

(a) The slit separation is $11.1 \mu m$.

(b) The total number of interference maxima is $50$
.

(c) The smallest angle for maxima is $0°$.

(d) The greatest angle for minima is $78.44°$ .

Step 1: Concept/Significance of diffraction

The expression for the slit width of the diffraction grating is given by,

 $$\begin{gathered} \mathit{\beta }\mathbf{=}\frac{\mathbf{\pi }\mathbf{d}}{\mathbf{\lambda }}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \\ \mathit{d}\mathbf{=}\mathbf{\left(}\frac{\mathbf{\beta }}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}\mathbf{\right)}\mathbf{\left(}\frac{\mathbf{\lambda }}{\mathbf{\pi }}\mathbf{\right)} \end{gathered}$$                                                                                              …… (1)

Here, d is slit separation, $\mathit{\beta }$ is slit width, $\mathit{\lambda }$ is wavelength, and $\mathit{\theta }$ is angle of diffraction.

Step 2: (a) Find the slit separation

From the graph, the ratio of slit with to sine of angle is,

 $\begin{array}{rcl}\frac{\beta }{\sin \theta }& =& \frac{80 \text{rad}}{1}\\ & =& 80 \text{rad}\end{array}$

 

Substitute $80 \text{rad}$ for $\frac{\beta }{\sin \theta }$, and $435\times {10}^{-9} \text{m}$ for in equation (1).

 $\begin{array}{rcl}d& =& \left(80 \text{rad}\right)\left(\frac{435\times {10}^{-9} \text{m}}{\pi }\right)\\ & =& 11.1\times {10}^{-6} \text{m}\\ & =& 11.1 \mu m\end{array}$

 

Therefore, the slit separation is $11.1 \mu m$ .

Step 3: (b) Find the total number of interference maxima

The condition for the constructive interference maxima is given by,

$\begin{array}{rcl}d\sin {\theta }_{\max }& =& m_{\max }\lambda \\ m_{\max }& =& \frac{d\sin {\theta }_{\max }}{\lambda }\end{array}$                                                                                     …… (2)

Here, ${\theta }_{\max }$ is the maximum angle, and $m_{\max }$ is the number of maxima.

 

Substitute $11.1\times {10}^{-6} \text{m}$ for $d$ , $90°$ for ${\theta }_{\max }$, and $435\times {10}^{-9} \text{m}$ for $\lambda$ in equation (2).

 $\begin{array}{rcl}{m}_{\max }& =& \frac{\left(11.1\times {10}^{-6} \text{m}\right)\sin 90°}{435\times {10}^{-9} \text{m}}\\ & =& 25.52\\ & \approx & 25\end{array}$

 

The number of interference maxima are 25 and the central maxima is formed at 0 . So, the number of interference minima are 25 .

 

The total number of interference maxima and minima on both sides of the interference pattern is,

 $$\begin{gathered} n=25+25 \\ =50 \end{gathered}$$

 

Therefore, the total number of interference maxima on both sides of the interference pattern is 50.

Step 4: (c) Find the smallest angle for a maxima

The least maxima is occurred at the center if the interference pattern.

 $m_{\max }=0$

 

From equation (1),

$$\begin{gathered} {\theta }_{\max }={\sin }^{-1}\left(\frac{m_{\max }\lambda }{d}\right) \\ ={\sin }^{-1}\left(\frac{\left(0\right)\lambda }{d}\right) \\ =0° \end{gathered}$$ 

 

Therefore, the smallest angle for maxima is $0°$ .

Step 5: (d) Find the greatest angle for a minimum

The number of minima is 25. So, the smallest minima is occurred at 1 and the greatest minima is occurred at 25 .

$m_{\min }=25$ 

 

The condition for interference minima is given by,

 $$\begin{gathered} d\sin {\theta }_{\min }=m_{\min }\lambda \\ {\theta }_{\min }={\sin }^{-1}\left(\frac{m_{\min }\lambda }{d}\right) \end{gathered}$$                                                                                  …… (3)

 

Substitute $11.1\times {10}^{-6} \text{m}$ for $d$ , $25$ for $m_{\min }$ , and $435\times {10}^{-9} \text{m}$ for $\lambda$ in equation (3).

$$\begin{gathered} {\theta }_{\min }={\sin }^{-1}\left(\frac{\left(25\right)\left(435\times {10}^{-9} \text{m}\right)}{11.1\times {10}^{-6} \text{m}}\right) \\ =78.44° \end{gathered}$$

 

Therefore, the greatest angle for minima is $78.44°$ .