Shift of fringes when a slab/lens in introduced in front of a slit
Example: n transparent slabs of refractive index 1.5 each, having thickness 1cm,2cm,.......... to ncm are arranged one over another. A point object is seen through this combination with the perpendicular light. If the shift of object by the combination is 1cm then find the value of n.
Solution: As the refractive index of all the slabs are 1.5 so those may be a considered as a single slab with the thickness summed up. Summed up thickness of n slabs Δt=1+2+3+...+n=2n(n+1) Given, for small angle of incidence, shift of object: s=(1−μ1)Δt so, ⇒1cm=(1−1.51)2n(n+1) ⇒6=n(n+1) ⇒n2+n−6=0 ⇒n2+3n−2n−6=0 ⇒n(n+3)−2(n+3)=0 ⇒(n+3)(n−2)=0 So, n=−3 or n=2 As, n can not be negative. so n=2
Define and calculate fringe width and angular fringe width in YDSE
Fringe width is the distance between two successive bright fringes or two successive dark fringes. In the interference pattern, the fringe width is constant for all the fringes. Fringe width is independent of order of fringe. Fringe width is directly proportional to wavelength of the light used. It is given by: β=dλD Angular fringe width is given by: tanθ≈θ=Dβ=dλ
Example: In Young's double slit experiment the two slits are illuminated by light of wavelength 5890∘A and the distance between the fringes obtained on the screen is 0.2∘. If the whole apparatus is immersed in water then find the angular fringe width. (refractive index of water is 4/3)
Solution: For interference in YDSE: dsinθ=nλ For small θ, sinθ can be approximated to θ So, dθ=nλ or, θ∝λ Now as the set up immersed in water, λ will become μλ=4/3λ=43λ So, λ2λ1=θ2θ1 or, 3λ/4λ=θ20.2∘ or, θ2=0.2∘×43=0.15∘.
Find the ratio of maximum to minimum intensity
Example: In Young's double slit experiment, first slit has width four times the width of the second slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe system.
Let intensity from the smaller slit be I. So, intensity from the bigger slit will be 4I since it is 4 times of smaller slit. Now, Imax=4I+I+2(4I2)=5I+24I2=5I+2(2I)=9I Imin=4I+I−2I(4I)=5I−2(4I2)=I. So, IminImax=I9I=19.