WBUT CS Solved Question Papers
Interference Of Light
The central fringe in a Newton’s ring experiment with a monochromatic light is
a) bright b) dark c) white [WBUT 2005(DECEMBER)]
1.1. The intensities of the two waves are in the ratio of 9 : 1 produce interference. The ratio of maximum to minimum intensity is equal to
a) 10: 8 b) 9 : 1 c) 4 : 1 d) 2 : 1 [WBUT 2006(DECEMBER)]
1.2. Interference fringes in Newton’s rings are example of
a) equal inclination, coherence is obtained by amplitude division
b) ‘* qual thickness, coherence is obtained by amplitude division
c) equal thickness, coherence is obtained by wave-front division
d) equal inclination, coherence is obtained by wave-front division Answer: (b)
1.3. A source of light emits lights of frequencies between v to v + Av. Coherence time of the emergent light beam is Tc, then [WBUT 2008(JUNE)]
a) Tc °c Av b) Tc 1/Av c) Tc d) Tc «= 1/v
1.4. Newton’s ring experiment is based on [WBUT 2009(JUNE)] a) division of amplitude b) division of wave-front – c) none of these
1.5. In an interference pattern produced by the identical coherent sources of monochromatic light, the intensity at the site of central maximum is I. If the
intensity at the same spot when either of the two slits is closed is 70, we must
have [WBUT 2010(JUNE)]
a) / = /0 b) / = 2/0 c) I—4I0 d) / and /0 are not related
1.6. Two sources are said to be coherent when the waves produced by them are
a) same wavelength
b) same wavelength and same phase
c) same wavelength and constant phase difference
d) same amplitude and constant phase difference [WBUT 2010(DECEMBER)]
1.7. In an arrangement for viewing Newton’s ring, if the lens which rests on a glass plate were moved upwards by one wavelength, (of the viewing light), which of the following will be observed? [WBUT 2010(DECEMBER)]
a) The central spot becomes bright b) No change of fringe pattern is observed
c) The rings shift towards the centre d) The rings move out from the centre Answer: (b)
1.8. In Newton’s ring experiment the diameters of the dark rings are proportional to
a) odd natural number [WBUT 2011 (DECEMBER)]
b) square root of the natural numbers
c) square root of the odd natural numbers
d) square of the natural numbers Answer: (b)
2.1. In a Young’s double slit interference experiment distance between the coherent sources are 1.15 mm. Calculate the fringe width that would be observed on a screen placed at a distance of 85 cm from the source. The wavelength of light used is 5893 A. [WBUT 2008(JUNE), 2010(JUNE)]
From the expression of fringe width we have
2.2. Interference phenomenon does not violate the principle of conservation of energy. Justify it. [WBUT 2008(JUNE)]
We know that the resultant intensity of the interference is I = 4a2 cos2 8/2 From the above equation it is clear that the intensity at bright points is 4a2 and at dark points it is zero. According to the law of conservation of energy, the energy cannot be destroyed. Here also the energy is not destroyed but only transferred from the points of minimum intensity to the points of maximum intensity.
For bright points, the intensity due to the two waves should
be 2a but actually it is 4a . As shown in the figure
intensity varies from 0 to 4a2, and the average is still 2a2. It 8 >
is equal to the uniform intensity 2a’ which will be present in the absence of the interference phenomenon due to the two waves. Therefore, the formation of interference fringes is in accordance with the law of conservation of energy.
2.3. A film of oil of r.i. 1.70 is placed between a plane glass late and an equi-convex lens. The focal length of the lens is 1 metre. Determine the radius of the 10th dark ring when light of wavelength is 6000 A. [WBUT 2008(JUNE), 2010(JUNE)]
The expression for focal length of a convex lens is — = (// — l)
convex lens /?, = R2 = R (say). So the expression of focal length will be
. We also know that radius of the nth dark ring
} 2 is represented by r= nXR —> rn2 = nA (// — 1) — (on substituting the value of R).
So the radius of the 10th dark ring will be ;;2 = 10x6000x10″8 x( 1.70 — 1) ^
100 or, r2 =840×10“* ->r■ =28.89×10^ cm
2.4. do you mean by temporal and spatial coherence? Define coherent length.
Let us assume a bundle of light waves propagating in space, as shown in the diagram. If the phase difference measured at a single point in the space at the beginning and end of a fixed time interval does not change with time, then the waves are said to have temporal coherence. The phase difference between any two fixed points along any ray will be independent of time. So it is also known as longitudinal coherence. Temporal coherence is characterized by two parameters viz. coherent length 1^ and coherent time xCoh* Coherent length or coherent time is a measure of how long (by distance or by time respectively) the wave maintains its phase when it is propagating in space.
wavelength, and AX is the band width of the packet. It is expressed in terms of frequency
as Lh —where v is the line width. So from the above mathematical expression it is
obvious that larger the bandwidth, coherent length decreases. A strictly monochromatic wave should be an ideal harmonic wave of infinite extent as Av is equal to zero. A natural light wave consists of a train of wave packets because of which its temporal coherence decreases. So mono-chromaticity is a measure of its temporal coherence.
Spatial coherence refers to the continuity and uniformity of a wave in a direction perpendicular to the direction of propagation. It is also known as lateral or transverse coherence. If the phase difference for any two fixed points in a plane normal to the wave propagation does not vary with time, then the wave is said to exhibit spatial coherence.
2.5. If the amplitudes of two coherent light waves are in the ratio 1 : 4, find the ratio of maximum and minimum intensity in the interference pattern.[WBUT 2009(JUNE)] Answer:
Intensity expression is given by, / = al +a2 + 2axa2 cos o
When 8 = 0 / = Imax = a2 + a2 + 2a,a2 = (a, + a2 )2
When S = 7T I = lmin=af+a2+2ala2=(al—a2)2 As per the problem,
Imax _ (flt +g2f _ (4 +1)2 _ 25 U («,-«2)2 (4-l)2 9
2.6. In a Newton’s ring experiment, the diameter of a dark ring is 0.32 cm, when the wavelength of monochromatic light be 6000 A. What would be the diameter of that ring when the wavelength of light changes to 5000 A. [WBUT 2010(DECEMBER)] Answer:
D = yj4mRA, where the symbols have their usual meaning
Now, 0.32 = V4mR x 6000x 10-8 = 77.4&j4mR Dm = y/4 mRA = yj4mR x5000xl0“8 D.„ V5000 f5 D ^0.32xV5 _Q29
2.7. In Young’s double slit experiment the distance between two slits is 0.5 mm. the wavelength of light is 5000 A and the separation between the sources and the screen is 50 cm. Calculate the Fringe width in this case. [WBUT 2010(DECEMBER)]
Using the relation (5 ——- we have [where the symbols have their usual meaning]
a DA 50x5000x10-® _n in_3 AA<
B =—- =————– :– cm = 50×10 cm = 0.05 cm.
Which is the required fringe width.
2.8. A mica sheet of thickness t and refractive index fj is introduced in the path of one of the interfering beams in Young’s double slit experiment. Find out the linear displacement of the nth bright fringes in terms of and ‘f. [WBUT 2011 (JUNE)]
Let t be thickness of the mica sheet c be the velocity of light in free space v be the
velocity of light through mica.
, ,. . .BP AP-t t From the diagram shown we can write,—————————————- =—————————————- h —
c c v
As time taken by the wave from B to P in air is the same as the time taken by the wave from A to P partly through air and partly through the plate.
=> BP = AP-t + -t
But —-ju Hence BP—AP=jut—t=(/l—l)t
If P is the point originally occupied by the nth fringe, then the path difference, BP— AP=nA or, (//—l)t=nA
Also the distance x through which the fringe is shifted =nj3 due to insertion of the mica
sheet of refractive index n . The value of n be integer.
a nAD . . 0 nAD . x=np —» x—————- ( on substituting p —- )
or, xd / D = nA xd / D=(ju—l)t
So the required fringe shift will be jc =
2.9. How are coherent sources produced in Young’s double slit experiment? Show that the law of conservation of energy is not violated in case of interference.
[WBUT 2011 (DECEMBER)]
The light wave diffracts and illuminates two narrow, parallel slits
5, and S2, which are
very close to each other.
The slits 5, and S2
partition the incident wavefront . In accordance with Huygens principle, two segments of the wavefront at slits
5, and S2 can be treated as point sources of
waves. These sources are generated by the same primary wave and are therefore mutually coherent. The waves that spread from 5, and S2 partially overlap. The overlapping waves interfere and produce interference fringes on the screen as shown.
We know that the resultant intensity of the interference is /=4a2cos2—
a) Coherence means the coordinated motion of several waves maintaining a fixed and predictable phase relationship with each other. Normally we divide coherence in two category stated below. (For the concluding part Refer to Question No.2.4.)
b) Refer to Question No. 2.2.
c) Given D = 100 cm.
d = .01 cm.
X = 5893A= 5893x 10 8 cm then using the formula
,aD A , U
x =—— and putting n= 1 we have,
As jj. is greater for liquid (oil) so by putting the value of \x in the above expression we get the radius of the bright ring will be decreased.
3.2. Newton’s ring arrangement is used with a source emitting two wavelengths A] = 600 nm and A^ = 590 nm. It was found that the nth dark ring due to A^
coincides with the (/i + l)th dark ring due to A^. If radius of curvature of the lens is 0.9 m then find out the value of n and the diameter of the nth dark ring due to A7. [WBUT 2011 (JUNE)]
Using the formula of Newton’s ring, we have the condition for n* dark ring D~ = AnAR ….(1)
where the symbols have their usual meaning . According to the problem wc can write 4nA,R = 4(n + \)A2R
or, –,l—=-?-■ Putting the values of A. =600nm and A, = 590nm we have —-— = — n +1 \ n ^ n + l 60
or, n = 59.
Putting n + l = 60 and ^=590 in D„+l=4(n + l)A2R we have D59=1.12cm
3.3. a) Distinguish between interference and diffraction. Is energy conserved in interference phenomenon? Explain.
b) In a Newton’s ring arrangement, the diameter of the 5th dark ring is 0.3 cm and the diameter of 25th dark ring is 0.8cm. If the radius of curvature of the planoconvex lens is 100 cm, find the wavelength of the light used.
c) What are the conditions to get sustained interference pattern?
|i) Superposition of two light rays coming from two different wave front originating from the same source is the basic of interference.||i) Superposition of light rays coming from different parts of the same wave front is the basic of diffraction phenomenon.|
|ii) Fringe width may or may not be same.||ii) Fringe width are not same.|
|iii) Intensity of all bright fringes are equal.||iii) Intensity of bright bands are not same.|
|iv) Minimum intensity are perfectly dark.||iv) Minimum intensity are not perfectly dark.|
|v) Intensity distribution of interference is uniform.||v) Non uniform distribution of intensity of diffraction pattern.|
2nrJ Part: Refer to Question No. 3.1(b).
„ D2_ -D2
b) Using the formula A = —£2–2——- — we have
where Dm= 0.3 (5th ring dia ) Dm+n= 0.8 (25th ring dia) n=20; Given #=100cm
A = —^ ——– LQ’.r^ = 6875×10~8cm = 6875 A which is the required wavelength.
c) The required conditions are:
- The two sources should be coherent (if the phase difference between the two waves emitted from two sources is zero or has a constant value with time, then the two sources are known as coherent sources. If the phase difference changes with time, the two sources are know as incoherent sources).
- The two sources must emit monochromatic wavelength having continuous waves and constant time period.
- The amplitudes of the two interfering beams should be equal or very nearly equal. The emitted waves should be preferably of same amplitude to get completely dark fringes.
- The separation between the two sources (2d) should be small and the distance D
between two sources and screen should be large to observe distinct fringes.
For constructive interference (i.e., for maximum intensity) the path difference
between the waves should be even multiple of —. For destructive interference (i.e., for minimum intensity) the path difference between the waves should be odd multiple
The two light waves must be in the same state of polarization.