In a typical Young's double slit experiment, S 1 and S 2 are identical slits and equidistant from a point monochromatic source S of light having wavelength λ . The distance between slits is represented by d and that between slits and screen is represented by D. P is a fixed point on the screen at a distance
y = 
from central order bright on the screen: where D o , d o are initial values of D and d respectively. In each statement of column- I some changes are made to above mentioned situation. The distance between the slits and the source is very large. The effect of corresponding changes is given in column- II . Match the statements in column- I with resulting changes in column- II .
Column- I column- II
. (neglecting absorption in medium)
Text Solution
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r,s p,q,s p,q,s r,s
Sol. The y-coordinate of point P = 
= half the fringe width. Hence initial intensity at P is zero.
As d is doubled, fringe width becomes half. As a result first order bright is formed at P and optical path difference between interfering waves at P will increase.
As D is doubled, fringe width doubles, hence intensity at P cannot be zero. Also optical path difference between interfering waves at P will decrease.
Decreasing width of slit will reduce energy from S 1 . But as D is doubled intensity at P is changed from zero to non zero value. Hence all options are same as in case B.
As set up is submerged in water of refractive index 
, fringe width reduces, optical path difference will become 
rd its initial value. Hence dark cannot be again obtained at P. Therefore intensity will increase P.
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