In a YDSE experiment, the distance between the slits & the screen is 100 cm. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width 0.25 mm. When the distance between the slits is increased by Δ d = 1.2 mm, the fringe width decreased to
n = 2/3 of the original value. In the final position, a thin glass plate of refractive index 1.5 is kept in front of one of the slits & the shift of central maximum is observed to be 20 fringe width. Find the thickness of the plate & wavelength of the incident light.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
λ = 600 nm , t = 24 μ m
Sol Clearly β initial = 
⇒ 0.25 × 10 –3 = 
× 1 m
⇒ 
= 2.5 × 10 –4 .............(1)
Afterwards
= 
⇒ 
= 
.............(2)
Dividing (1) and (2)
= 
⇒ d = 2 ( Δ d) = 2.4 mm.
& λ = 2.4 × 2.5 × 10 –7 m = 600 nm.
Now 
Now P becomes central maxima.
⇒ for point P d sin θ = ( μ – 1) t
⇒ d. 
tan θ ≈ ( μ – 1)t
⇒ d = ( μ – 1)t
⇒ 
= (1.5 – 1)t
⇒ t = 
= 
= 24 μ m.
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