A monochromatic light of λ = 5000 Å is incident on two slits separated by a distance of
5 x 10 − 4 m. The interference pattern is seen on a screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10 − 6 m & refractive index μ = 1.5 is placed between one of the slits & the screen. Find the intensity at the centre of the screen, if the intensity there is Ι 0 in the absence of the plate. Also find the lateral shift of the central maximum.
Text Solution
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( 0 , 1.5 mm)
Sol. 
After the thin plate is placed between one of the slits and the screen, an additional path difference [ Δ P = ( μ – 1) t ] is created between the rays emerging out of S 1 and S 2 . Thus
S 1 O – S 2 O = ( μ – 1) t = 0.75 × 10 –6 m
= 
⇒ ( μ – 1)t = 
Point O would be an interference minimum. Thus intensity at O is zero.
S 2 P – S 2 P = 
S 2 P – S 1 P = 
– ( μ – 1)t
For central maxima, the path difference between the interfering beams is zero.
i.e. S 2 P – S 1 P = 0
⇒ 
= ( μ – 1)t
⇒ y m = 
( μ – 1)t = 
×0.75 × 10 –6 = 1.5 mm
Thus, the effect of putting a thin film in front of one of the slits is to shift the central maximum towards that slit.
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