Team Quanta gladly presents all possible short questions of BS Physics book Waves and Oscillation Chapter#05: Diffraction and Interference.
Q.1 What changes occur in the pattern of interference fringes if double slit apparatus is placed under water?
Answer: Width of bright or dark fringe is given by,
Fringe width =λD/d
When double slit apparatus is dipped in water wavelength λ decreases so width of dark and bright fringes decreases.
Q.2 What causes the fluttering of TV picture when air-plane flies overhead?
Answer: TV signals from TV station, reaching our TV
- Directly
- After reflection from air-plane interfere at TV set so fluttering of TV picture takes place.
Q.3 Is it possible to have coherence but when light sources emitting light of different wavelengths?
Answer: No, different sources emitting light of different wavelengths are not phase coherent. Phase coherent sources are derived from one signal source and light rays emitted from these sources are of some wavelength
Q.4 What is meant by dispersion ‘D’ of grating?
Answer: A measurement of the ability of a substance to repartee by refraction expressed by the first differentiate of the refractive index with respect to wavelength at a given value of wavelength ‘D’.
Q.5 What is importance of Bragg’s law?
Answer: The Bragg law is useful for measuring wavelengths and for determining the lattice spacings of crystals. To measure a particular wavelength, the radiation beam and the detector are both set at some arbitrary angle θ. The angle is then modified until a strong signal is received.
Q.6 Why head light beam of an automobile does not produce interference when overlap at distant point?
Answer: In the case of car headlights, the two sources are independent and have no coherence. The phase of light emitted by one headlight world have a random relation with the light from the other headlight, and there will be no interference.
Q.7 For transmitted light, why the central spot of Newton’s ring is bright?
Answer: The two rays transmitted from a thin film of thickness L are in phase and so interfere constructively, since one ray has to travel an intra distance of 2L (for normal incidence ) than the other ray. Therefore, the condition for observing intensity maximum is 2L =and the central spot of Newton’s rings bright.
Q.8 Why do coated lenses look purple by reflected light?
Answer: The purpose of coating on a lens is to reduce reflections from the lens surface. Light reflects from the surface of the glass because the refractive index of the glass is different from that of the air, and the reflection is formed at this interface of different refractive indices. This reflected wave has a cancellation effect on combining with the incident light thus reducing the light intensity and hence the coated lenses look purple.
Q.9 Light has wavelength frequency, and speed which any one of these quantities remains unchanged when light passes from a vacuum into a slab of glass?
Answer: Frequency remains unchanged. Speed decreases due to decrease in wavelength.
Q.10 The speed and wavelength of, say red light that we see in air are reduced when light passes into water. Would that light then appear to be another color blue perhaps if you view it from the underwater surface?
Answer: No as frequency of light remains unchanged while passing through water so its color also remains unchanged.
Q.11 Why are parallel slits are preferable to pinholes that Young used in demonstrating interference?
Answer: Due to long narrow slits, bright and dark bands of equal width are obtained at the screen.
Q.12 Describe the pattern of light intensity on screen if one slit is covered with a red filter and other with a blue filter, the incident light being white.
Answer: The light rays emitted from red and blue filters will become non-phase coherent, so interference pattern on screen will disappear.
Q.13 A person wets his eyeglasses to clean them. As the water evaporates he notices that for a short time, the glasses become markedly less reflecting. Explain why?
Answer: As the water evaporates, salts and impurities present in water remains deposited on the glass. This thin layer of salts and impurities absorbs the light falling on glasses so reflection from glasses decreases.
Q.14 A lens is coated to reduce reflection what happens to the energy that had previously been reflected. Is it absorbed by the coating?
Answer: The energy now is partly refracted and partly absorbed by coating and glass. Energy and intensity reduce.
Q.15 An automobile directs its head lights onto the side of a Barn (rough building), why are interference fringes not produced in the region in which light from the two beams overlap.
Answer: Beam from two head –lights are not phase coherent. So they do not interfere in the region where they overlap.
Q.16 A soap film on a wire loop held in air appears black at its thinnest portion when viewed by reflected light. On the other hand, a thin oil film floating on water appears bright at its thinnest portion when similarly viewed from the air above explain.
Answer: At the top of bubble soap film thickness d of film is zero, so apparently path difference between two rays one reflected from upper side of film and other from lower side appears zero, but as rays reflected from upper side is reflected from denser medium so a path difference of N2 is produced. Hence actual path difference is
$$\ s=2d-\lambda/2\ \ \ \ \ \$$
$$s=\ 2\ x\ 0\ -\ \lambda/2$$
$$s=\ -\ \lambda/2$$
As path difference is so interference is destructive and top of film appears dark. In horizontal oil film on water, both waves, from top and bottom of oil film are reflected from denser medium so path difference λ is produced in both waves. Hence actual path difference becomes λ/2 at the thinnest part of film. Therefore constructive interference takes place and film appears bright.
Q.17 Why does a film (soap bubble) have to be thin to display interference effects? Or does it? How thin in “thin” discuss.
Answer: A white light have seven colors. When white light falls on the soap bubble this film have two sides one is upper and the other is lower than the two reflected rays produce interference effect. The thickness of soap bubbles is very small i.e. in micron so it is called thin film.
Q.18 The central spot of Newton’s ring is black. Why?
Answer: Where the two glass surfaces are in direct contact, the thick of air film is zero thus the path difference between the reflected and refracted rays is also zero. The refracted ray which reflects off the plane glass undergoes a 180° phase shift, so the two rays cancel each other via destructive interference. Thus a dark spot is formed at the center.
Q.19 If the path length in the movable mirror in Michelson’s interferometer greatly exceeds that to fixed mirror (say by more than a meter), the fringes begin to disappear. Explain why, lasers greatly extends this range. Why?
Answer: To see interference fringes, the property “coherence of length” must be ensured. If the differences in path length to both the mirrors and back from the beam splitter is greater than the coherence length, then no interference effects will be seen. On the other hand, the long coherence length of laser light, because of its highly directional nature, makes it easy to obtain interference fringes and interferometer path lengths no longer have to be matched as they do If a short coherence length white light source is used.
