BABU BANARASI DAS NORTHERN INDIA INSTITUTE OF TECHNOLOGY,

LUCKNOW

B. TECH. FIRST SEMESTER 2014-2015

SUBJECT- PHYSICS

UNIT III : INTERFERENCE

SHORT QUESTIONS WITH ANSWERS

Q.1 Define coherent sources. Discuss why two independent sources of light of same wavelength

cannot show interference?

A.1 Two sources of light are said to be coherent if they vibrate in same phase or there exists a constant

phase relationship. Two independent sources of light of same wavelength cannot show interference

because they cannot maintain the requirement of constant phase difference between them.

Q.2 State two methods of producing coherent sources of light?

A.2 Two methods of producing coherent sources of light are by

(I)Division of wavefront –Fresnel’s Biprism

(ii) Division of amplitude-Newton’s ring

Q.3 What change will you observe in the Young’s double slit experiment pattern if the whole

arrangement is dipped in water?

A.3 If the whole arrangement is dipped in water in Young’s double slit experiment, and then the fringe

width decreases and the fringes come closer due to the change of the refractive index of the medium.

Q.4 State the essential conditions for observing the phenomenon of interference of light.

A.4 The essential conditions for observing the phenomenon of interference of light are as follows

Two sources should be coherent, monochromatic and having nearly same amplitude.

Q. 5 Explain the colours, when a thin film illuminated by white light is observed in reflected light.

A .5 When a thin film is illuminated by white light beautiful colours will be observed. These colours arise

due to the interference of light waves reflected from the upper and lower surfaces of the film.

.

Q.6. Why an extended source is necessary to observe colours in thin film?

A.6 If we employ an extended source, the light reflected by every point of the film reaches the eye.

Hence the entire film can be viewed simultaneously by keeping the eye at one place only.

Q.7 Explain why an excessively thin film appears black and excessively thick film shows no colour

in reflected light?

A.7. For an excessively thin film the effective path difference in reflected light (2µtcos r + λ/2) becomes

λ/2, as2µtcos r is negligible in comparison to λ/2.This is the condition of minimum intensity for all

wavelengths. Therefore all wavelengths will be absent in reflected system, hence film will appear dark.

Q.8. Explain why an excessively thick film shows no colour in reflected light?

A.8. The bright and dark appearance of reflected light depends upon the values of µ,t and r. In case of

white light even if t and r made constant µ varies with wavelength. At a given point or for any value of r,

due to large thickness (t)a large number of wavelengths can be found to satisfy the condition of

constructive interference for every colour of the spectrum and on the other hand, at the same point

condition of destructive interference is also satisfied for another set of wavelengths. Moreover number of

wavelengths sending the maximum intensity is almost equal to the number of wavelengths sending the

minimum intensity. Also, fringes due to colours satisfying the condition of maximum intensity will

superimpose and produce uniform illumination.

Q.9 Why Newton’s rings are circular but air wedge fringes are straight?

A.9 In both Newton’s rings and air wedge fringe arrangements, each fringe is the locus of points of equal

thickness of the film. In Newton’s rings, the locus of points of equal thickness of air film lie on a circle

with the point of contact as centre. Hence fringes are circular and concentric. In case of air wedge, the

locus of points of equal thickness is straight lines parallel to the edge of the wedge. Hence the fringes are

straight and parallel.

Q.10 Why the center of Newton’s rings appears dark in reflected light?

A.10 In case of Newton’s rings the effective path difference between the interfering rays is

2µtcos(r+α)+λ/2. At the centre t=0 and for a very small angle of wedge cos(r+α) =1.Therefore effective

path difference at the point of contact = 2µt+λ/2. At the center t=0, hence effective path difference is λ/2.

This is the condition of minimum intensity. Hence central spot of the ring system appears dark.

Q.11 What will happen to Newton’s rings if a drop of water is introduced between the glass plate

and lens?

A.11 Since diameter of the ring is inversely proportional to the square root of refractive index of the

medium between the planoconvex lens and plane glass plate, therefore as refractive index increases,

diameter of the rings decrease, that is the rings contract.

Q.12 What will happen to Newton’s rings if the lens is lifted up slowly from the flat surface?

A.12 As the distance between the lens and the plate is increased or the lens is lifted up slowly from the

flat surface, the order of the ring at a given point increases. The rings, therefore, come closer and closer

until they can no longer be separately observed.

Q.13 What will happen to Newton’s rings if monochromatic light is replaced by white light?

A.13 If monochromatic light is replaced by white light in Newton’s rings experiment, a few coloured

rings around a black centre are observed and beyond it a uniform illumination is obtained. This is because

the diameter of the ring is directly proportional to square root of wavelength as and white light is

composed of a number of colours, diameter of different rings will be different. In ring pattern violet ring

will be the innermost and red ring will be the outermost and due to overlapping of rings of different

colours only first few rings will be clearly visible.

Q.14 What will happen to Newton’s rings if the plane glass plate is replaced by plane mirror?

A.14 If the plane glass plate is replaced by plane mirror in Newton’s rings experiment, the ring on the

reflected system would disappear and a uniform illumination is observed. It is because there would be no

transmission of rays, but transmitted rays will also be reflected at the silvered surface and the two

complementary systems of rings superimpose on each other and give a uniform illumination.

Q.15 What will happen to Newton’s rings if a lens of small radii of curvature is used?

A.15 As diameters of rings (bright or dark) are directly proportional to the square root of the radius of

curvature, R of the lens. Hence, the spacing between two consecutive rings goes on decreasing as we

move towards the centre of lens.