question_answer

 A screen is at a distance \[D=80cm\]from a diaphragm having two narrow slits \[{{S}_{1}}\]and \[{{S}_{2}}\]which are\[d=2mm\]apart. Slit \[{{S}_{1}}\] is covered by a transparent sheet of thickness \[{{t}_{1}}=2.5\mu m\]and \[{{S}_{2}}\]by another sheet of thickness \[{{t}_{2}}=1.25\mu m\]as shown in figure. Both sheets are made of same material having refractive index\[\mu =1.40\]. Water is filled in space between diaphragm and screen. A monodichromatic light beam of wavelength \[\lambda =5000~\overset{\text{o}}{\mathop{\text{A}}}\,\] is incident normally on the diaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio of intensity at C to maximum intensity of interference pattern obtained on the screen, where C is foot of perpendicular bisector of \[{{S}_{1}}{{S}_{2}}.\] (Refractive index of water,\[{{\mu }_{W}}=4/3\]). If it is\[\frac{a}{4}\]. Find\[a\].

Answer:

3.00