n12 (b) 5 7
3 11 5 22
n n
n n
+ + (c) 3 5
5 11 7 22
n n
n n
+
+ (d) 7 3
5 12 3 21
n n
n n
+ +
55. If the value of g at the surface of the earth is 9.8 m s–2, then the value of g at a place 480 km above the surface of the earth will be
(Radius of the earth is 6400 km) (a) 8.5 m s–2 (b) 9.8 m s–2 (c) 7.2 m s–2 (d) 4.2 m s–2 56. Resultant of two vectors
Aa dn B is of magnitude P.
If
B is reversed, then resultant is of magnitude Q.
What is the value of P2 + Q2?
(a) 2(A2 + B2) (b) 2(A2 – B2) (c) A2 – B2 (d) A2 + B2
57. The speed of an electromagnetic wave in a material medium of permeability m and permittivity e is (a) 1me (b) 1
2me (c) 1
me (d) 1 2me 58. Magnetic field at the centre of a circular loop of area
A is B. The magnetic moment of the loop will be (a) BA2
m p0 (b) BA3 2
0 /
m p (c) BA3 2
0 1 2 /
m p/ (d) 2 3 2
0 1 2
BA/ m p/
59. The system as shown in figure is released from rest with the spring initially stretched 75 mm.
Calculate the velocity v of the block after it has dropped 12 mm.
The spring has a spring constant 1050 N m–1. Neglect the mass of the pulley.
(a) 0.271 m s–1 (b) 0.371 m s–1 (c) 2.71 m s–1 (d) 3.71 m s–1
60. Effective capacitance between points A and B as shown in the figure.
Here C1 = C2 = 20 mF and C3 = C4 = 10 mF
(a) 10 mF (b) 15 mF (c) 20 mF (d) 25 mF
ANSWER Keys
1. (c) 2. (d) 3. (d) 4. (b) 5. (d) 6. (c) 7. (d) 8. (b) 9. (b) 10. (a) 11. (b) 12. (a) 13. (d) 14. (a) 15. (b) 16. (b) 17. (d) 18. (d) 19. (b) 20. (d) 21. (a) 22. (b) 23. (a) 24. (b) 25. (c) 26. (c) 27. (c) 28. (c) 29. (b) 30. (c) 31. (c) 32. (b) 33. (a) 34. (d) 35. (b) 36. (d) 37. (a) 38. (a) 39. (b) 40. (a) 41. (a) 42. (d) 43. (c) 44. (d) 45. (a) 46. (c) 47. (c) 48. (d) 49. (b) 50. (a) 51. (b) 52. (b) 53. (d) 54. (b) 55. (a) 56. (a) 57. (c) 58. (d) 59. (b) 60. (b)
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Solution of April 2016 CroSSWord
We begin a complicated topic with a simple question.
What do you expect when a parallel beam of monochromatic light is made to strike a slit (whose dimension is comparable to wavelength) on a screen?
The answer appears obvious, i.e. through the slit light will pass through while from the remaining portion it would be blocked.
But experimental results give contradictory results.
Reality
Screen Slit
Screen
Light Alternate illuminated
(bright fringes) and dark regions with the intensity of bright fringes decreasing as we move away from the central bright fringe.
}
This experimental result suggests that light spread beyond the slit too, i.e., light bends on striking the slit to reach regions which would have been impossible to reach, had it travelled in a straight line.
The alternate bright and dark fringes suggest that interference is also taking place. But unlike interference, diffraction is a complicated phenomenon since here we have infinite number of point sources within the slit from which light waves are reaching on the screen.
Side view of the experimental arrangement shows this:
The intensity pattern shows that central bright fringe is the most intense and wide fringe while for secondary bright fringes the intensity decreases so largely that the bright fringes become almost undetectable beyond 3 to 4 fringes on either side of central maxima.
We can mathematically calculate the approximate locations of these fringes by using the concept of half period zones (HPZs).
Each HPZ is a zone on the slit from the top and bottom of which if we consider two rays, there will be a path difference of l/2 (half period) between them.
To find out the locations of
1. Dark fringes - we divide the slit width into even number of HPZs.
2. Bright fringes - we divide the slit width into odd number of HPZs.
Why do we do so will be answered gradually in the sections ahead.
Location of 1st dark fringe
We divide the slit width into 2 HPZs.
Considering slit width = b
\ b = 2 HPZs
Contributed By: Bishwajit Barnwal, Aakash Institute, Kolkata
Now the path difference between the two rays shown in the figure is l/2 (definition of HPZ).
P b/2
b/2 x x
1 32 4
\ In pair, since the path difference between ray 1 and 2 is l
2, therefore for every ray 3 chosen between ray 1 and 2, there will be another ray 4 with which it will have a path difference of l/2. Hence we can see that the waves coming out from one HPZ undergoes destructive interference with another HPZ beneath it.
Therefore, at the location P on the screen we get the first dark fringe for which the condition would be
∆x b= d = ⇒ b d=
2sinq1 l2 sinq1 l If q1d is very small, then sinq1d ≈ q1d
\ ⋅b = ⇒ b y =
d Dd
q1 l 1 l⇒ y = D
d b
1 l
Till this location from the centre of the screen on either side, we would get a gradually decreasing intensity, with maximum intensity at the centre.
Hence the width of central maxima is
W y W D
c=2 d ⇒ c=2b
1 l
Angular fringe width of the central maxima is the angular separation between the two extremities of the central maxima as seen
from the slit.
\ If Wc << D, then
q l
c Wc
D b
= = 2
Location of 2nd dark fringe Slit width = b = 4 HPZs
\ Here in pair, 1st HPZ undergoes destructive interference with the 2nd and 3rd with the 4th. So finally we don't have any light source (secondary wavelets) whose waves stand without undergoing destructive interference with any other. So we get a dark fringe.
From diagram,
∆x b= d=
4sinq2 2l (definition of HPZ)
\ bsin q2d = 2l
\ Now we generalise and say, that for the location of nth dark fringe,
Slit width = 2n HPZs
\ b = ⇒ =
n nd b nd n
2 sinq 2l sinq l
where qnd is angular position of the point on the screen (where nth dark fringe is formed) measured with respect to centre of the slit.
Similarly, for location of bright fringes we divide the slit width into odd number of HPZs. This is done since, one HPZ undergoes destructive interference with the other while there would be one HPZ left over at last which will stand alone. Hence the intensity will be non-zero on the screen due to it. One more advantage is that as we move towards higher order maxima, the width of that single HPZ left over goes on decreasing due to which we have less number of secondary sources whose intensity will be seen for higher order maxima which explains the decrease in the intensity for higher order maxima.
Ist secondary maxima (bright fringe) Slit width = 3 HPZs
\ From diagram,
∆x b= B= 3sinq1 2l
⇒ bsinqB= l
1 3
\ Generalising, for nth order secondary maxima,2 slit width = (2n + 1) HPZs
\ b+ =
n nB
2 1sinq 2l ⇒ bsinqnB=(2 1n+ )l 2
Note that the results for location of bright and dark fringes are just the reverse of the results of Young's double slit experiment (YDSE).
Another striking difference between YDSE and diffraction through single slit is that we get large number of fringes of almost equal intensity in YDSE whereas in diffraction beyond 3rd to 4th order bright fringe the intensity decreases to such a large extent that it almost becomes invisible.
Hence, the angular position q of the bright/dark fringes in YDSE is generally small for which sinq q≈ ≈ y
D can easily be used but in diffraction the angular position in numericals are large, such as 30° for which sinq ≠ q.
Hence be careful on this front while solving numericals.
Diffraction through circular aperture If instead of slit, we have a
Screen small hole of dimensions
comparable to wavelength then the waves get diffracted from the entire periphery of the circle and hence the fringes formed on the
screen are circular and alternately bright and dark with gradually decreasing intensity of the bright fringes.
The central bright fringe is the most intense bright fringe and is known as airy disc.
From side view if we redraw the set up we can see :
Unlike diffraction through single slit, here the first minima is formed at angular position,
q=1 22l≈ 2 . ( )a
r D
Images formed by lens are also diffraction images, since the aperture of the lens is circular. Hence if two point objects are placed very close to each other, their airy disc image might overlap with each other and hence we would not be able to resolve (differentiate) them.
Rayleigh lays a criterion for resolvability and says that if the airy disc images formed are such that the centre of one airy disc coincides with the first minima, i.e., the boundary of the other airy disc, they would be said to be just resolved. For better clarity let us try to understand this with a pictorial representation.
d
O1 O2
I2
I1
The expected images are point images I1 and I2 for point objects O1 and O2 but due to diffraction from boundary of lens, airy disc images are formed and when viewed different situations may arise as below:
I2
I1
I2 I1
I2 I1
Clearly resolved images (since they
don't overlap)
Even though slightly overlapped but is considered to be
resolved
Just resolved (According to
Rayleigh)
If we go on decreasing the separation between the objects, separation between airy disc image also decreases and hence this puts a limit on how close we can place so that they can be resolved.
Two point objects separated by an angle q are said to be resolved if
q≥1 22. l d ,
where d is diameter of aperture of lens.
The equality hold for just being resolved.
Limit of resolution of a telescope or human eye is defined as the smallest angular separation at which two point objects may be kept such that they can be just resolved (according to Rayleigh criterion).
\ q = l
min 1 22.
where d = diameter of objective lens in telescope or d diameter of eye lens in human eye.
Note that for human eye limit of resolution is approximately
′ =
° = ×
1 1
60 1 1
60 180 π rad
Resolving power of any optical instrument is just the inverse of limit of resolution. Hence greater the limit of resolution, lesser is the resolving power.
nn
paper-2
1. (4) : Potential difference across the capacitor is q (for normal incidence). But there is abrupt change of l/2 in path of light at upper surface. So actual path difference is 2 µt – l/2
For constructive interference, 2 µt – l 2 = nl field at an angle q = 60° with the original direction.
X Total volume of wasted water = p 6 10
2 distance r from centre of earth O. Since, the particle is constrained to move along the tunnel, we define its position as distance x from C. Hence, equation of motion of the particle is,
max = Fx
The gravitational force on mass m at distance r is, F = GMmr
Contd. from page no. 21