91Ó°ÊÓ

The separation distance between double slit arrangement is the 100 times the light passes through the slits,$d=100\mathrm{λ}$.

The condition for the maxima in Young’s experiment is given as follows.$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}\mathbf{=}\mathit{m}\mathit{λ}$ …… (1)

Here, $\mathit{λ}$. is the wavelength, mis the order and$\mathit{θ}$is the angular separation.

The expression to calculate the distance between the central axis and first minima is given as follows.

$\mathit{y}\mathbf{=}\mathit{L}\mathit{s}\mathit{i}\mathit{n}\mathit{θ}$..…. (2)

Here,$\mathit{L}$ is the distance between the slit and screen.

a.

Calculate the angular separation in radians between the central maximum and an adjacent maximum as below.

Substitute 1 for m and $100\mathrm{λ}$ for d into equation (1).

$$\begin{gathered} \left(100\mathrm{λ}\right)\sin \mathrm{θ}=1×\mathrm{λ} \\ \sin \mathrm{θ}=\frac{\mathrm{λ}}{100\mathrm{λ}} \end{gathered}$$

$$\begin{gathered} \mathrm{θ}={\sin }^{-1}\left(\frac{1}{100}\right) \\ =0.572° \end{gathered}$$

Convert the angular separation from degrees to radian.

$$\begin{gathered} \mathrm{θ}=0.572×\frac{\mathrm{π}}{180} \\ =0.01\text{rad} \end{gathered}$$

Hence the angular separation between the central maximum and adjacent maximum is 0.01 rad.

b.

The distance between slit and screen, $L=50\text{cm}$

Calculate the distance between central axis and first maximum.

Substitute 50 cm for L and $0.{572}^0$for $\mathrm{θ}$into equation (2)

$$\begin{gathered} y=50×{10}^{-2}\sin \left(0.572°\right) \\ =0.50×0.01 \\ =0.50\text{cm} \end{gathered}$$

Hence, the distance between central axis and first maximum is 0.50 cm.