A) fringe width will decrease
B) fringes will disappear
C) fringe width will increase
D) there will be no change in fringe width.
Correct Answer: D
Solution :
Suppose \[{{S}_{1}}\]and \[{{S}_{2}}\] are the slits at a distance d from each other. Distance of screen be D. Let P be a point where there is a bright fringe. A glass plate is placed in the path of the ray from \[{{S}_{1}}\] to P. We know that the path difference between the rays in absence of glass plate is \[\Delta x={{S}_{2}}P-{{S}_{1}}P=\frac{dy}{D}\] In presence of the glass plate, the optical path length of the ray from \[{{S}_{1}}\] to P will be different. The total optical path length for this ray is given by \[{{S}_{1}}P-t+\mu t\] \[={{S}_{1}}P+(\mu -1)t\] Where \[\mu \] is the refractive index of the glass plate and is its thickness. Hence the new path difference is given by \[\Delta x={{S}_{2}}P-[{{S}_{1}}P+(\nu -1)t]\] \[=\Delta x-(\mu -1)t\] \[=\frac{dv}{D}-(\mu -1)t\] For a bright fringe. \[\Delta x=n\lambda \] and \[y=yn=\] distance of the bright fringe from the central fringe \[\therefore \] \[\frac{d{{y}_{n}}}{D}-(\mu -1)t=n\lambda \] \[\Rightarrow \] \[{{y}_{n}}=\frac{D}{d}[(\mu -1)t+n\lambda ]\] \[\therefore \] \[{{y}_{n+1}}-{{y}_{n}}=\omega =\frac{D\lambda }{d}\] Hence the fringe width remains constant.
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