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A light wave that has vibrations in more than one plane is known as unpolarized light.
Notes:
Concept:
Brewster angle
The angle of incidence at which a beam of unpolarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarising or Brewster angle. It is denoted by ip.
The British Physicist David Brewster found the relationship between Brewster angle (ip) and refractive index (μ) –
μ = tan ip
This relation is known as Brewster law.
Calculation:
Given:
Critical angle,
ic=sin−1(35)∴sinic=35
As μ=1sinic=53
According to Brewster’s law
tan ip = μ
Where ip is the polarising angle
∴ tanip=53ip=tan−1(53)
CONCEPT:
EXPLANATION:
CALCULATION:
Given: Angular width of the central maximum = 2θ
θ = π/6 = 30° Width of the slit (b) = 1.2 μm = 1.2 × 10-6 m
We know, b sin θ = n λ
For central maxima n = 1
⇒ b sin θ = λ
λ=1.2×10−6×sin30∘=1.2×10−6×12=6×10−7m=6000×10−10m = 6000 Å
Intensity ∝ (Amplitude)2
I1I2=(A1A2)2
The maximum intensity of interference:
Imax=(A1+A2)2=(I1+I2)2
Minimum intensity of the interference
Imin=(A1−A2)2=(I1−I2)2
I1I2=(A1A2)2=21
⇒A1A2=I1I2=91=31
⇒A1=3A2
ImaxImin=(A1+A2)2(A1−A2)2=(A1A2+1)2(A1A2−1)2
=(3+1)2(3−1)2=164=41
∆x = n λ, n = 0, 1, 2....
λ is the wavelength of light, n is an integer representing the number of the particular interference.
Δx=(n+12)λ,n=0,1,2...
We have to find find the path difference ∆x when the second minima (dark fringe) is formed.
putting n = 1 in the formula of the dark fringe
Δx=(1+12)λ
Δx=(32)λ
So, 3 λ/2 is the correct option.
Important Points
Young's double-slit Experiment: The Young's double-slit Experiment was done by Thomas Young in 1851.
Young's double-slit experiment
Fringe width (β):
⇒β=λDd
Where d = distance between slits, D = distance between slits and screen, and λ = wavelength
Given λ = 580 nm = 580 × 10-9 m, d = 0.30 mm = 0.30 × 10-3 m, and D = 2m
⇒β=λDd -----(1)
So the width of the central fringe is given as,
⇒ W = 2β
⇒W=2λDd
⇒W=2×580×10−9×20.3×10−3
⇒ W = 7.7 × 10–3 m
EXTRA POINTS:
The following are the conditions for the interference to happen:
Sl. No
Interference
Diffraction
1
Results due to superposition of waves from two coherent source
Results due to superposition of wavelets from different parts of same wave-fronts (single coherent source)
2
All fringes are of the same width β = λ D/d
All secondary fringes are of the same width but the central maximum is of double the width β0 = 2β = 2λ D/d
3
All fringes are of the same intensity
Intensity decreases as the order of maximum increases
Additional Information
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