Physics Question on Wave optics

Question

In a Fresnel biprism experiment, the two positions of lens give separation between the slits as 16 cm and 9 cm, respectively. What is the actual distance of separation?

Answer 1

Solution

Fresnel used a biprism to obtain two coherent sources for producing interference fringes in the laboratory. It consists of two acute angled prisms with their bases in contact. In actual practice, the biprism is constructed as a single prisms of obtuse angle of about 179 $^{\circ}$ and the remaining two acute angles are 30' each. In bi prism, the virtual images act as two coherent superimposed and interference fringes are formed in overlapping region AB on a screen placed at 0.

In order to measure the distanced between the virtual sources $S_1$ and $S_2$ , Glazebrook gave a method, known as magnification method due to Glazebrook. If $d_1$ and $d_2$ are distance between real images of $S_1$ and $S_2$ , then $d = \sqrt{d_1 d_2}$
Given $d_1 = 16 \, cm, \, \, d_2 = 9 \, cm$
$\therefore d = \sqrt{16 \times 9} = 12 \, cm$