Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Displacement of Fringes When a film of thickness’t’ and refractive index 'm' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes. Optical path difference at P = S2P - [S1P+ µt - t] = S2P - S1P - (µ - 1)t = y.d/D - (µ - 1) ⇒ nth fringe is shifted by Δy = D(µ-1)t/d = w/λ (µ-1)t Illustration: Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10-5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet? Solution: As derived earlier, the total fringe shift = w/λ (µ-1)t . As each fringe width = w, The number of fringes that will shift = total fringe shift/fring width (w/λ(µ-1)t)/w = (µ-1)t/λ = (1.6-1) x 1.8 x 10-5m / 600 x 10-9 = 18 Illustration: In the YDSE conducted with white light (4000Å-7000Å), consider two points P1 and P2 on the screen at y1=0.2mm and y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points. Solution: The optical path difference at P1 is p1 =dy1/d = ( 10/4000) (0.2) = 5 x 10-4mm = 5000Å In the visible range 4000 - 7000Å n1 = 5000/4000 = 1.25 and n2 = 5000/7000 = 0.714 The only integer between 0.714 and 1.25 is 1 ∴ The wavelength which forms maxima at P is l = 5000Å For the point P2, p2 = dy2/D = ( 10/4000) (1.6) = 4 x 10-3mm = 40000Å Here n1 = 40000/4000 = 10 and n2 = 40000/7000 = 5.71 The integers between 5.71 and 10 are 6, 7, 8, 9 and 10 ∴The wavelengths which form maxima at P2 are λ1 = 4000 Å for n = 10 λ2 = 4444 Å for n = 9 λ3 = 5000 Å for n = 8 λ4 = 5714 Å for n = 7 λ5 = 6666 Å for n = 6
When a film of thickness’t’ and refractive index 'm' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes.
Optical path difference at
P = S2P - [S1P+ µt - t] = S2P - S1P - (µ - 1)t = y.d/D - (µ - 1)
Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10-5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet?
As derived earlier, the total fringe shift = w/λ (µ-1)t .
As each fringe width = w,
The number of fringes that will shift = total fringe shift/fring width
(w/λ(µ-1)t)/w = (µ-1)t/λ = (1.6-1) x 1.8 x 10-5m / 600 x 10-9 = 18
In the YDSE conducted with white light (4000Å-7000Å), consider two points P1 and P2 on the screen at y1=0.2mm and y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points.
The optical path difference at P1 is
p1 =dy1/d = ( 10/4000) (0.2) = 5 x 10-4mm = 5000Å
In the visible range 4000 - 7000Å
n1 = 5000/4000 = 1.25 and n2 = 5000/7000 = 0.714
The only integer between 0.714 and 1.25 is 1
∴ The wavelength which forms maxima at P is l = 5000Å
For the point P2,
p2 = dy2/D = ( 10/4000) (1.6) = 4 x 10-3mm = 40000Å
Here n1 = 40000/4000 = 10 and n2 = 40000/7000 = 5.71
The integers between 5.71 and 10 are 6, 7, 8, 9 and 10
∴The wavelengths which form maxima at P2 are
λ1 = 4000 Å for n = 10
λ2 = 4444 Å for n = 9
λ3 = 5000 Å for n = 8
λ4 = 5714 Å for n = 7
λ5 = 6666 Å for n = 6
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Diffraction Table of Content Diffraction Pattern...
Young’s Double Slit Experiment Table of...
Polarisation Table of Content Polarisation of...
Questions and Solutions part 3 1. When light is...
Interference by Thin Film Table of Content Colours...
Wave Nature of Light Table of Content Huygens Wave...
Questions and Solutions part 2 11. Angular width...
Questions and Solutions Part 1 1. In Young's...
