91Ó°ÊÓ

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

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

The angle, $\mathrm{θ}=20°$.

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

$$\begin{gathered} \mathit{ϕ}\mathbf{=}\frac{\mathbf{1}}{\mathbf{λ}}\mathit{Δ}\mathit{L} \\ \mathit{ϕ}\mathbf{=}\frac{\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{θ}}{\mathbf{λ}} \end{gathered}$$…… (1)

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

${\mathit{ϕ}}_{\mathbf{rad}}\mathbf{=}\mathbf{2}\mathit{π}\mathit{ϕ}$…… (2)

a.

Calculate the multiple of $\mathrm{λ}$.

Substitute $4.24\mathrm{μ}\text{m}$ for d, $20.0^0$for $\mathrm{θ}$and 500 nm for $\mathrm{λ}$into equation (i).

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

Hence, the multiple of $\mathrm{λ}$ which gives the phase difference between the two rays is $2.9$.

b.

Calculate the phase difference in radian.

Substitute 2.9 for $âˆ\dots$into equation (2).

$$\begin{gathered} {\mathrm{ϕ}}_{rad}=2\mathrm{π}×2.9 \\ =5.8×3.14 \\ =18.22\text{rad} \end{gathered}$$.

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

c.

The phase difference for the bright fringes is given by,

$$\begin{gathered} {\mathrm{ϕ}}_{bright}=\frac{d\sin \mathrm{θ}}{\mathrm{λ}} \\ =m_{bright} \end{gathered}$$

The phase difference for the dark fringes is given by,

${\mathrm{ϕ}}_{dark}=\frac{d\sin \mathrm{θ}}{\mathrm{λ}}$

${\mathrm{ϕ}}_{dark}=m_{dark}+\frac{1}{2}$ …… (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.