Question

In the two-slit experiment of Fig.35-10, let angle $\mathit{\theta }$be ${\mathbf{20}\mathbf{.}\mathbf{0}}^{\mathbf{0}}\mathit{C}$, the slit separation be $\mathbf{4}\mathbf{.}\mathbf{24}\mathbf{}\mathit{\mu }\mathit{m}$, and the wavelength be $\mathit{\lambda }\mathbf{=}\mathbf{500}\mathbf{}\mathbf{\text{nm}}$. (a) What multiple of $\mathit{\lambda }$gives the phase difference between the waves of rays ${\mathit{r}}_{\mathbf{1}}$and ${\mathit{r}}_{\mathbf{2}}$when they arrive at point $\mathit{P}$on the distant screen? (b) What is the phase difference in radians? (c) Determine where in the interference pattern point $\mathit{P}$ lies by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies?

Short answer

a. The multiple of $\mathrm{\lambda }$which gives the phase difference between the two rays is2.9.

b. The phase difference in radian is $18.22\text{rad}$.

c. The point $P$lies second minimum and third maximum.

Step-by-step solution

 Write the given data from the question

The slit separation, $d=4.24\mu \text{m}$.

The wavelength, $\lambda =500\text{nm}$.

The angle, $\theta =20°$.

Determine the formulas to calculate the multiple of λ, phase difference and position of point P lies.

The expression to calculate the phase difference between the two rays is given as follows.

$$\begin{gathered} \mathit{\varphi }\mathbf{=}\frac{\mathbf{1}}{\mathbf{\lambda }}\mathit{\Delta }\mathit{L} \\ \mathit{\varphi }\mathbf{=}\frac{\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{\lambda }} \end{gathered}$$…… (1)

The expression to calculate the phase difference in radian is given as follows.

${\mathit{\varphi }}_{\mathbf{rad}}\mathbf{=}\mathbf{2}\mathit{\pi }\mathit{\varphi }$…… (2)

Calculate the multiple of λ which gives the phase difference between the two rays

a.

Calculate the multiple of $\lambda$.

Substitute $4.24\mu \text{m}$ for d, $20.0^0$for $\theta$and 500 nm for $\lambda$into equation (i).

$$\begin{gathered} \varphi =\frac{4.24\times {10}^{-6}\sin \left(20°\right)}{500\times {10}^{-9}} \\ =\frac{1.45\times {10}^{-6}}{500\times {10}^{-9}} \\ =0.0029\times {10}^3 \\ =2.9 \end{gathered}$$

Hence, the multiple of $\lambda$ which gives the phase difference between the two rays is $2.9$.

Calculate the phase difference in radian

b.

Calculate the phase difference in radian.

Substitute 2.9 for $\varnothing$into equation (2).

$$\begin{gathered} {\varphi }_{rad}=2\pi \times 2.9 \\ =5.8\times 3.14 \\ =18.22\text{rad} \end{gathered}$$.

Hence, the phase difference in radian is $18.22\text{rad}$.

Calculate the position of the point P in the interference

c.

The phase difference for the bright fringes is given by,

$$\begin{gathered} {\varphi }_{bright}=\frac{d\sin \theta }{\lambda } \\ =m_{bright} \end{gathered}$$

The phase difference for the dark fringes is given by,

${\varphi }_{dark}=\frac{d\sin \theta }{\lambda }$

${\varphi }_{dark}=m_{dark}+\frac{1}{2}$ …… (3)

The phase difference 2.9 is less than 3 which would respond to third side maximum and greater than 2.5 which would respond to second minimum and when the $m_{dark}=2$then by the equation (3) the phase difference would be 2.5.

Hence, the point P lies second minimum and third maximum.