Definitions [7]
A fringe is a bright or dark band formed on the screen due to constructive or destructive interference.
Interference is the phenomenon in which light intensity is modified due to the superposition of two or more light waves.
Fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes.
The condition where the images of two point objects close to each other are regarded as resolved (separated), if the central maximum of one falls on the first minimum of the other, is called Rayleigh's criterion for resolution.
When the separation between the central maxima of two objects is less than the distance between the central maximum and the first minimum of any of the two objects, the images are said to be 'not resolved' or unresolved.
When the separation between the central maxima of the two objects is just equal to the distance between the central maximum and first minimum of any of the two objects, the images are said to be just resolved.
When the separation between the central maxima of two objects is greater than the distance between the central maximum and first minimum of any of the two objects, the images are said to be well resolved.
Formulae [7]
\[\beta=\frac{\lambda D}{d}\]
\[\alpha=\frac{\beta}{D}=\frac{\lambda}{d}\]
\[x_n-x_m=\left[n-\frac{(2m-1)}{2}\right]\beta\]
\[\begin{array} {cc} & x_m=\frac{(2m-1)\lambda D}{2d} \end{array}\]
\[\begin{array} {c}x_n=\frac{n\lambda D}{d}=n\beta \end{array}\]
Limit of resolution (self-luminous objects):
d = \[\frac{1.22\lambda}{2\sin\theta}=\frac{0.61\lambda}{\sin\theta}\]
Limit of resolution (objects illuminated by light of wavelength λ):
d = \[\frac{\lambda}{2\sin\theta}\]
If a liquid of refractive index μμ is between object and objective:
d = \[\frac{\lambda}{2\mu\sin\theta}\]
where θθ is the angle subtended by an object at the objective.
Smallest angular separation dθdθ (Circular aperture):
\[d\theta=\frac{1.22\lambda}{D}\]
where D is the aperture (diameter) of objective of the telescope.
Smallest angular separation (Rectangular aperture):
\[d\theta=\frac{d}{D}\]
where d is slit separation and D is distance.
Theorems and Laws [1]
Statement: According to Lord Rayleigh, the images of two point objects close to each other are regarded as resolved (separated), if the central maximum of one falls on the first minimum of the other.
The reasoning of Rayleigh's criterion is given by considering the intensity distribution in the diffraction pattern produced by two objects.
Three Conditions:
| Condition | Description |
|---|---|
| Not Resolved | Separation between central maxima < distance between central maximum and first minimum of either object |
| Just Resolved | Separation between central maxima = distance between central maximum and first minimum of either object |
| Well Resolved | Separation between central maxima > distance between central maximum and first minimum of either object |
Important Questions [34]
- The Co-efficient of Reflection of an Opaque Body is 0.16. Its Co-efficient of Emission is
- Distinguish between the phenomenon of interference and diffraction of light.
- State the Conditions to Get Steady Interference Pattern.
- State the Conditions to Get Constructive and Destructive Interference of Light.
- In Interference Pattern, Using Two Coherent Sources of Light; the Fringe Width is ________
- In a Biprism Experiment, the Light of Wavelength 5200å is Used to Get an Interference Pattern on the Screen. the Fringe Width Changes by 1.3 Mm When the Screen is Moved Towards Biprism by 50 Cm. Find the Distance Between Two Virtual Images of the Slit.
- A parallel beam of monochromatic light is incident on a glass slab at an angle of incidence 60o. Find the ratio of width of the beam in the glass to that in the air if refractive index of glass is 3/2.
- A Parallel Beam of Light Travelling in Water is Incident Obliquely on a Glass Surface. After Refraction Its Width ..............
- Two Coherent Sources of Light Having Intensity Ratio 81 : 1 Produce Interference Fringes. Calculate the Ratio of Intensities at the Maxima and Minima in the Interference Pattern.
- In Young’S Experiment, the Ratio of Intensity at the Maxima and Minima in an Interference Pattern is 36 : 9. What Will Be the Ratio of the Intensities of Two Interfering Waves?
- The Fringes Produced in Diffraction Pattern Are of .
- Using Analytical Method for Interference Bands, Obtain an Expression for Path Difference Between Two Light Waves.
- In Young' S Experiment the Ratio of Intensity at the Maxima and Minima . in the Interference Pattern is 36 : 16. What is the Ratio of the Widths of the Two Slits?
- In Young’S Experiment Interference Bands Were Produced on a Screen Placed at 150 Cm from Two Slits, 0.15 Mm Apart and Illuminated by the Light of Wavelength 6500 å. Calculate the Fringe Width.
- Derive an Expression for Path Difference in Young’S Double Slit Experiment
- Using Monochromatic Light of Wavelength λ in Young’S Double Slit Experiment, the Eleventh Dark Fringe is Obtained on the Screen for a Phase Difference of
- In Biprism Experiment, 10th Dark Band is Observed at 2.09 mm from the Central Bright Point on the Screen with Red Light of Wavelength 6400 A.
- In a Biprism Experiment, When a Convex Lens Was Placed Between the Biprism and Eyepiece at a Distance of 30 Cm from the Slit, the Virtual Images of the Slits Are Found to Be Separated by 7 Mm.
- In a Biprism Experiment, a Slit is Illuminated by a Light of Wavelength 4800 A The Distance Between the Slit and Biprism is 15 Cm and the Distance Between the Biprism and Eyepiece is 85 Cm.
- A Red Light of Wavelength 6400a° in Air Has Wavelength 4000a° in Glass. If the Wavelength of Violet Light in Air is 4400a°, Find Its Wavelength in Glass.
- The Width of Plane Incident Wavefront is Found to Be Doubled on Refraction in Denser Medium. If It Makes an Angle of 65° Medium. with the Normal, Calculate the Refractive Index for the Denser Medium.
- Describe Biprism Experiment to Calculate the Wavelength of a Monochormatic Light. Draw the Necessary Ray Diagram.
- In Biprism Experiment Two Interfering Waves Are Produced Due to Division of
- Describe Biprism Experiment to Find the Wavelength of Monochromatic Light. Draw the Necessary Ray Diagram for Magnified and Diminished Images of Virtual Sources.
- Monochromatic Light of Wavelength 4300 å Falls on a Slit of Width ‘A’. for What Value of ‘A’ the First Maximum Falls at 30° ?
- Calculate Width of the Slit and Width of the Central Maximum.
- In the Diffraction Pattern Due to a Single Slit of Width 'd' With Incident Light of Wavelength 'λ', at an Angle of Diffraction θ. the Condition for First Minimum is
- Find The· Nature of Illumination at the Point If Wavelength of Light is 4000a.
- Explain Two Types Of Diffraction of Light.
- What is 'Diffraction of Light'
- If 'A' is the Aperture of Telescope and A. is the Wavelength of Light Then Resolving Power of Telescope is .............
- The Resolving Power of Telescope of Aperture 100 Cm for Light of Wavelength 5.5 X 10^-7 m
- The Numerical Aperture of Objective of a Microscope is 0.12. the Limit of Resolution, When Light of Wavelength 6000A is Used to View an Object is
- A Point is Situated at 6.5 Em and 6.65 Em from Two Coherent Sources. Find the Nature of Illumination at the Point, If Wavelength of Light is 5000 å
Concepts [9]
- Interference of Light
- Conditions for Producing Steady Interference Pattern
- Interference of Light Waves and Young’s Experiment
- Analytical Treatment of Interference Bands
- Measurement of Wavelength by Biprism Experiment
- Fraunhofer Diffraction Due to a Single Slit
- Rayleigh’s Criterion
- Resolving Power of a Microscope and Telescope
- Difference Between Interference and Diffraction
