In a YDSE experiment, the two slits are covered with transparent membranes of negligible thickness which allow light to pass through it but does not allow water. A glass slab of thickness t = 0.41 mm and refractive index μ g = 1.5 is placed in front of one of the slits as shown in figure. The separation between the slits d = 0.30 mm. The entire space to the left of the slits is filled with water of refractive index μ w = 
. A coherent light of intensity I and absolute wavelength λ = 5000Å is being incident on the slits making an angle of 30° with horizontal. Screen is placed at a distance D = 1 m from the slits.
(i) At point O, equidistant from slits we get –
Text Solution
Verified by Experts(i) Sol; (ii) Sol; (iii) Sol
Ans.
(i)
Sol. At O, optical path diff.,
Δ x = ( μ g – 1)t – μ w d sin θ
⇒ Δ x = 0.5 × 0.41 × 10 –3 – 
× 3 × 10 –4 × 
= 5 × 10 – 6 m
= 10 λ
i.e. 10 th bright fringe.
(ii)
Sol. For central maxima,
Δ x = 
+ ( μ g – 1)t – μ w d sin θ = 0
⇒ y 
= – 10 λ
⇒ y = – 10 
= 
× 10 –2 m
(iii)
Sol. At P,
Δ x = 
+ ( μ g – 1) t – μ w d sin θ
= 
+ 10 λ = 10 λ + 
Δφ = 
× 
= 
I P = 
= I 
I max = 2I ∴ 
= 
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