The fringe width ($\beta$) in Young's double-slit experiment is determined by the wavelength of light ($\lambda$), the distance between the slits and the screen ($D$), and the separation between the slits ($d$). The formula is:
$ \beta = \frac{\lambda D}{d} $
The problem states that two separate experimental set-ups produce fringes of equal width. Let the parameters for the two set-ups be subscripted with 1 and 2.
Therefore, we have:
$ \beta_1 = \beta_2 $
Substituting the formula for fringe width:
$ \frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2} $
We need to find the ratio of the distances between the slits and the screens, which is $D_1/D_2$. Rearranging the equation from the previous step:
$ \frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2} $
The problem provides the following ratios:
Substitute these values into the rearranged equation:
$ \frac{D_1}{D_2} = \left(\frac{2}{1}\right) \times \left(\frac{2}{1}\right) $
$ \frac{D_1}{D_2} = \frac{4}{1} $
The corresponding ratio of the distances ($D_1/D_2$) is 4.
As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface.
If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is.