Question

Describe the biprism experiment to find the wavelength of the monochromatic light. Draw the necessary ray diagram. The width of plane incident wavefront is found to be doubled on refraction in denser medium.

Answer 1

Used a biprism to get interference pattern by division of wavefront method. The biprism consist of two active angled prisms with their bases in contact. Here two sources $S_1$ and $S_2$ are the virtual images of the fine slit S as shown in Figure. The experimental arrangement consist of a slit S the biprism ABC and the microscope $M$. All are mounted on an optical bench. These are adjusted at the same height and can move and rotate as required. The light emerging from the slit fall on the biprism. The edge A of the biprism divides the incident wave front into two parts. One is through upper half AB of biprism and appears to coming from virtual source $S_1$. Other is from lower half AC of biprism and appears to coming virtual $S_2$. The interference fringes are seen in the overlapping region XY and can be seen by the eyepiece.

Adjustment of Bi-prism

To obtain well defined interference fringes following adjustments of apparatus are required.The optical bench should be leveled.

1. Slit, biprism and eyepiece should be adjusted to the same height.
2. Eyepiece should be adjusted to make a fine image at cross wires.
3. The slit should be narrow and vertical.
4. Bi-prism is positioned in such away, maybe by giving lateral movement that overlapping region should be in the field of view of the eyepiece.
5. The fringes so obtained should be made clear by adjusting biprism slit and eyepiece.
6. The lateral shift of the fringes should be removed with the help of biprism and eyepiece. The reason for the lateral shift is that the line joining ·the slit and biprism is not parallel to the length of the optical bench.
7. The bench should not be disturbed once these adjustments have been done. On the screen at O we get a central maximum then alternatively dark, bright fringes on both sides.

${Y_n}_{Bright}=n\lambda D$--------------(1)

${Y_n}_{drack}=\frac{(2n+1)\lambda D}{2d}$---------(2)

And fringe width is

$\beta =\frac{\lambda D}{d}$.

Now when the plane is incident in denser medium, the new wave front is

${\beta }^{\prime }=\frac{\beta }{\lambda (\mu -1)}$