Question Bank
Unit 2
Q.1 Explain total internal reflection?
Answer:
The phenomenon which occurs when the light rays travel from a more optically denser medium to a less optically denser medium.
Consider the following situation. A ray of light passes from a medium of water to that of air. Light ray will be refracted at the junction separating the two media. Since it passes from a medium of a higher refractive index to that having a lower refractive index, the refracted light ray bends away from the normal. At a specific angle of incidence, the incident ray of light is refracted in such a way that it passes along the surface of the water. This particular angle of incidence is called the critical angle. Here the angle of refraction is 90 degrees. When the angle of incidence is greater than the critical angle, the incident ray is reflected back to the medium. We call this phenomenon total internal reflection.
Notations Used in The Total Internal Reflection Formula and Critical Angle
- r is the angle of refraction
- i is the angle of incidence
- n1 is the refractive index in medium 1
- n2 is the refractive index in medium 2
- Ө is the critical angle
Following are the two conditions of total internal reflection:
- The light ray moves from a denser medium to less dense medium.
- The angle of incidence must be greater than the critical angle.
When the incident ray falls on the cladding, it suffers total internal reflection as the angle formed by the ray is greater than the critical angle. Optical fibres have revolutionised the speed with which signals are transferred, not only across cities but across countries and continents making telecommunication one of the fastest modes of information transfer. Optical fibres are also used in endoscopy.
Q 2 Explain Construction and working of Ruby laser?
Answer:
The ruby laser is the first type of laser actually constructed, first demonstrated in 1960 by T. H. Maiman. The ruby mineral (corundum) is aluminium oxide (Al2O3 ) with a small amount (about 0.05%) of chromium ions (Cr3+) which gives it its characteristic pink or red color by absorbing green and blue light.
The ruby laser is used as a pulsed laser, producing red light at 6943Å. After receiving a pumping flash from the flash tube, the laser light emerges for as long as the excited atoms persist in the ruby rod, which is typically about a millisecond.
The ruby laser has following main parts:
1. The working substance (active medium)- is in the form of a rod of ruby crystal (10 cm in length, 0.8 cm in diameter) in which Cr3+ are active centres while Al and O2 are inert.
2. The resonance cavity- is made by silvering and polishing the ends of ruby rod. Fully reflecting plates at the left and a partially reflecting plate at the right, both optically plane and accurately parallel.
3. The optical pumping system -consists of a helical xenon discharge tube. It produces flash of few milliseconds.
Ruby laser uses a three-level pumping scheme. The xenon discharge generates a flash of white light for few milliseconds. The Cr+3 ions are excited to the E3 level by absorbing green and blue component of white light. From there the Cr+3 ions undergo non-radiative transitions and quickly drop to the metastable level E2. The metastable state has greater life time than E3. Therefore Cr+3 ions accumulate at E2. When more than half ions are accumulated at E2 the population inversion is established between E2 and E1. A chance spontaneous emission leads to chain stimulated emission. Red light (of wavelength 6943 Å) emerges from the front face.
Q 3 what are Use of laser?
Answer:
1. Welding and Cutting: The highly collimated beam of a laser can be focused to a microscopic dot of extremely high energy density for welding and cutting. The automobile industry makes extensive use of carbon dioxide lasers with powers up to several kilowatts for computer-controlled welding on auto assembly lines.
2. Communication: The lasers have significant advantages in communication because they are more nearly monochromatic and this allows the pulse shape to be maintained better over long distances when used in optical fibre.
3. Barcode Scanner: Supermarket scanners typically use helium-neon lasers to scan the universal barcodes to identify products. The laser beam bounces off a rotating mirror and scans the code, sending a modulated beam to a light detector and then to a computer which has the product information stored.
4. Laser Printing: The laser printer has in a few years become the dominant mode of printing in offices. The laser is focused and scanned across a photoactive selenium coated drum where it produces a charge pattern which mirrors the material to be printed. This drum then holds the particles of the toner to transfer to paper which is rolled over the drum in the presence of heat.
5. CD’s and Optical Discs: The detection of the binary data stored in the form of pits on the compact disc is done with the use of a semiconductor laser.
6. Surveying and Ranging: Helium-neon and semiconductor lasers have become standard parts of the field surveyor's equipment. A fast laser pulse is sent to a corner reflector at the point to be measured and the time of reflection is measured to get the distance.
7. Laser cooling: The use of lasers to achieve extremely low temperatures has advanced to the point that temperatures of 10-9 K have been reached
8. Laser Spectroscopy: Laser spectroscopy has led to advances in the precision with which spectral line frequencies can be measured, and this has fundamental significance for our understanding of basic atomic processes.
9. Holography: Holography is "lens less photography" in which an image is captured not as an image focused on film, but as an interference pattern at the film.
Q 4 Define following?
A. Population inversion
B. Optical fibre
Answer:
Population inversion:
1. Energy is applied to active medium raising active centres to excited energy level.
2. These atoms spontaneously decay to a relatively long-lived, lower energy, metastable state
3. A population inversion is achieved when the majority of atoms have reached this metastable state
4. Lasing action occurs when an electron spontaneously returns to its ground state and produces a photon.
5. If the energy from this photon is of the precise wavelength, it will stimulate the production of another photon of the
Same wavelength and resulting in a cascading effect.
6. The highly reflective mirror and partially reflective mirror continue the reaction by directing photons back through the medium along the long axis of the laser.
7. The partially reflective mirror allows the transmission of a small amount of coherent radiation that we observe as the “beam”.
8. Laser radiation will continue as long as energy is applied to the lasing medium.
Optical fibre:
The optical fibre is a cylindrical, long, thin transparent structure made of glass and plastic, which is designed to guide the light wave from one end to another. The light inside the fibre is guided on the principle of Total Internal Reflection (TIR).
Optical fibres are widely used in fibre-optic communications to send information (data).
The basic structure of an optical fiber consists of three parts the core, the cladding, and the coating or buffer which are coaxially arranged. The innermost region is called the core, the light in the fibre travels only in the core. The core is surrounded by cladding, which is responsible for keeping the light inside the core. The refractive index of core (n1) is greater than of cladding (n2). The outermost region is called buffer or sheath, which protects the core and cladding from external abrasions
Q.5 Write some application of optical fibre?
Answer:
- Medical
Used as light guides, imaging tools and also as lasers for surgeries - Defence/Government
Used as hydrophones for seismic waves and SONAR , as wiring in aircraft, submarines and other vehicles and also for field networking - Data Storage
Used for data transmission - Telecommunications
Fiber is laid and used for transmitting and receiving purposes - Networking
Used to connect users and servers in a variety of network settings and help increase the speed and accuracy of data transmission - Industrial/Commercial
Used for imaging in hard to reach areas, as wiring where EMI is an issue, as sensory devices to make temperature, pressure and other measurements, and as wiring in automobiles and in industrial settings - Broadcast/CATV
Broadcast/cable companies are using fiber optic cables for wiring CATV, HDTV, internet, video on-demand and other applications
Fiber optic cables are used for lighting and imaging and as sensors to measure and monitor a vast array of variables. Fiber optic cables are also used in research and development and testing across all the above-mentioned industries.
Q 6Newton’s rings are observed in the reflected light of wave length 5900 Å. The diameter of 10th dark ring is 0.5 cm. Find the radius of curvature of the lens used.
Answer:
The given data are
Wave length of light (λ) = 5900 Å= 5900 × 10–10 m
Diameter of 10th Newton’s dark ring (D10) = 0.5 cm = 0.5 × 10–2 m
Radius of curvature of lens (R) =?
Formula is D2n = 4nλR
Q 7 Calculate the thickness of air film at the 10th dark ring in a Newton’s rings system, viewed normally by a reflected light of wave length 500 nm. The diameter of the 10th dark ring is 2 mm.
Answer:
Given data are
Wave length of light (λ) = 500 nm = 500 × 10–9 m
Number of the dark ring viewed (n) = 10
Diameter of 10th dark ring (D10) = 2 mm = 2 × 10–3 m
Radius of 10th dark ring 
Thickness of air film (t) =?
Condition for dark ring is
Q 8 What are condition of interference?
Answer:
Condition for Interference
To obtain the stable interference pattern the following conditions are required
Condition for Sustained Interference Pattern
a. The two interfering waves should be coherent i.e., both light waves are in same phase or maintain constant phase difference between them.
b. The source must be monochromatic
c. Both the wave must be in same set of polarization.
Condition for observation of fringes
a. The distance between the coherent sources be small
b. The distance between the source and screen must be large
Condition for Good Contrast
a. The amplitude of the both interfering waves be the same or very nearly same.
b. The sources must be narrow
c. The interfering angle should be very small or both travel in the same direction.
Q 9 Explain the Rayleigh criterion for limit of resolution?
Answer:
Rayleigh Criterion of Resolution
Statement: Two sources are resolvable by an optical instrument when the central maximum of one diffraction pattern falls over the first minimum of the other diffraction pattern and vice versa.
For example:
Let us consider the resolution of two wavelengths and by a grating. When the difference in wavelengths is smaller and such that the central maximum of the wavelength coincides with the first minimum of the other as shown in figure, then the resultant intensity curve is as shown by the thick curve. The curve shows a distinct dip in the middle of two central maxima. Thus, the two wavelengths can be distinguished from one another and according to Rayleigh they are said to be “Just Resolved”.
If the difference in wavelengths is such that their principal maxima are separately visible, then there is a distinct point of zero intensity in between the two wavelengths. Hence according to Rayleigh, they are said to be “Resolved”.
When the difference in wavelengths is so small that the central maxima corresponding to two wavelengths come still closer as shown in figure, then the resultant intensity curve is quite smooth without any dip. This curve is as if there is only one wavelength somewhat bigger and stronger.
Hence according to Rayleigh, the two wavelengths are “Not Resolved”.
Thus, the two spectral lines can be resolved only up to a certain limit expressed by Rayleigh Criterion.
Q 10Define Brewster angle?
Answer:
When unpolarized light of a certain wavelength is incident upon the surface of a transparent substance it experiences maximum plane polarization at the angle of incidence whose tangent is the refractive index of the substance.
The specific value of angle of incidence at which the reflected light is completely polarized is called the polarizing angle
Where,
- µ is the refractive index of the medium
- Ip is the polarizing angle
Brewster’s law states that, the tangent of polarising angle of incidence to the transparent medium is equal to the refractive index of the medium
Tanip=mu
Ip- polarising angle of incidence
Ip= tan–1 μ
The angle is depending on the refractive index of the medium.
