Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits S 1 and S 2 . In each of these cases S 1 P 0 = S 2 P 0 , S 1 P 1 – S 2 P 1 = λ /4 and S 1 P 2 – S 2 P 2 = λ /3, where λ is the wavelength of the light used. In the cases B, C and D, a transparent sheet of refractive index μ and thickness t is pasted on slit S 2 . The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by δ (P) and the intensity by Ι (P). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation. Column–I Column–II
(p) δ (P 0 ) = 0
(q) δ (P 1 ) = 0
(r) Ι (P 1 ) = 0
(s) Ι (P 0 ) > Ι (P 1 ) (t) Ι (P 2 ) > Ι (P 1 )
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
→ p, s; → q; → t ; → r, s, t
Sol. Ι (P 1 ) = Ι 1 + Ι 2 + 2 
cos 
= Ι 0 + Ι 0 + 2 Ι 0 . 
= (2 + 
) Ι 0
Ι (P 2 ) = Ι 1 + Ι 2 + 2 Ι 0 . 
= 2 
. cos 
= Ι 0 + Ι 0 + 2 Ι 0 . 
= Ι 0
∴ Ι (P 1 ) > Ι (P 2 )
δ (P 0 ) = ( μ – 1)t . 
= 
δ (P 1 ) = 
δ (P 2 ) = 
Ι (P 0 ) = Ι 1 + Ι 2 + 2 
. cos δ (P 0 )
= Ι 0 + Ι 0 + 2 Ι 0 . cos 
= 2 Ι 0
Ι (P 0 ) = 4 Ι 0
Ι (P 2 ) = Ι 0 + Ι 0 + 2 Ι 0 cos 
= (2 + 
) Ι 0
. δ (P 0 ) = ( μ – 1)t . 
= 
δ (P 1 ) = 
δ (P 2 ) = 
Ι (P 0 ) = 
δ (P 0 ) = 
Ι (P 1 ) = 0
Ι (P 2 ) > 0, Ι (P 0 ) > 0
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