JEE Main & Advanced Physics Wave Optics / तरंग प्रकाशिकी Fresnel's Biprims

(1) It is an optical device of producing interference of light Fresnel's biprism is made by joining base to base two thin prism of very small angle

(2) Acute angle of prism is about \[1/{{2}^{o}}\] and obtuse angle of prism is about \[{{179}^{o}}\].

(3) When a monochromatic light source is kept in front of biprism two coherent virtual source \[{{S}_{1}}\] and \[{{S}_{2}}\]are produced.

(4) Interference fringes are found on the screen placed behind the biprism interference fringes are formed in the limited region which can be observed with the help eye piece.

(5) Fringe width is measured by a micrometer attached to the eye piece. Fringes are of equal width and its value is \[\beta =\frac{\lambda \,D}{d}\]

(6) Let the separation between \[{{S}_{1}}\] and \[{{S}_{2}}\] be d and the distance of slits and the screen from the biprism be a and b respectively i.e. \[D=(a+b)\]. If angle of prism is \[\alpha \] and refractive index is \[\mu \] then \[d=2a(\mu -1)\alpha \] \[\therefore \]  \[\lambda =\frac{\beta \,[2a\,(\mu -1)\alpha ]}{(a+b)}\]\[\Rightarrow \]\[\beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }\]

(7) If a convex lens is mounted between the biprism and eye piece. There will be two positions of lens when the sharp images of coherent sources will be observed in the eyepiece. The separation of the images in the two positions are measured. Let these be d1 and d2 then \[d=\sqrt{{{d}_{1}}{{d}_{2}}}\]     \[\therefore \]\[\lambda =\frac{\beta d}{D}=\frac{\beta \sqrt{{{d}_{1}}{{d}_{2}}}}{(a+b)}\].