Questions and Answers of Engineering Physics

1. Describe with necessary theory the formation of interference pattern in an air wedge. Explain how it can be used to test the plainness of a glass plate. An air wedge is formed two glass plates of length 5cm and a thin wire. The wavelength of light used is 589 nm. If 200 fringes are formed in 1 mm, find the radius of the wire? (KU May 2011)             

Solution: Consider a wedge shaped liquid film formed between the surface of two plane glass plate. Let θ be the angle of wedge and μ be the refractive of the film.

   

For normal incidence, the path difference between the rays is given by

 ∆=2μ t cosr

When r=0, ∆=2μ t

 From fig, t=x tanθ

∴∆=2μ x tanθ

 The condition for brightness is

   2μ x tanθ= (2n-1)

 Condition for darkness is

  2μ x tanθ=n2μ

Where n=0,1,2,3⋯

 If x₁ is the distance of the nth dark band from the edge and x₂ that of the (n+m)^th dark, then

 

Therefore band width, β=

For air film, μ=1

∴β=

 The same relation holds good for bright bands also. The interference fringes obtained from a film of varying thickness have been used to test the optical planes of surfaces. If a wedge shaped air film is formed between an optically plane glass plate and the surface under test, the fringes will be straight if the surface is perfectly plane, otherwise irregular in shape.

1=5 cm

x=589 cm

β=1/200

2. What is mean by temporal and spatial coherence? (KU MAY 2010)

     Coherence ensures a constant phase difference at each point. Electric field associated with a wave at a point changes in phase with time. If the phase difference varies directly with time, then the wave is said to have temporal coherence. Spatial coherence refers to the coherence between the vibrations at different points on a plane perpendicular to the direction of propagation of the beam.

     The electric field (associated with a wave) at a point charges in phase with time. Let

     ∆ φ be the phase in a small interval of time ∆t. If ∆ φ is directly proportional to ∆t for any value of  v, then the wave is said to have temporal coherence.

     The beam is said to have perfect spatial coherence if the phase of vibration at any two points on such a plane differ by a constant value.                                                   

3. Explain Rayleigh’s criterion for resolution and write expression for resolving power of a telescope.   (KU MAY 2010)

Rayleigh’s criterion:

                        To measure the resolving power of an optical instrument, Lord Rayleigh suggested that two sources or their images can be regarded as separate (resolved) if central maxima of one pattern falls over the first minima of the other. This is equivalent to the condition that the distance between centers of patterns should be equal to the radius of the Airy’s disc or radius of the first dark ring. This condition is called Rayleigh’s criterion.

    Resolving power of a telescope is R=1/dθ = a/1.22

  Where ‘a’ is the diameter of the objective of the telescope.

4. Explain the test of planeness of surfaces?   (KU May 2009)

       If a wedge shaped air film is formed between an optically plane glass and the surface under test, the fringes will be straight the surface is perfectly plane; otherwise irregular in shape. The contour of a particular fringe would indicate the same thickness of the air film.

5. Define crystal lattice, basis and crystal structure?                                           (KU May 2008)

      Basis:- A basis in the simplest instance will be single atom and in a sophisticated crystal structure, it may consist of thousands of atoms of different elements.

      Crystal lattice: It is a mechanical abstraction used to represent the periodicity with which matter (atoms or molecules) is distributed in the crystal.

      Crystal structure  =Lattice+ basis

6. What are Miller indices? Explain how they are obtained. Obtain an expression for interplanar spacing of cubic crystal.                   (KU May 2007,2008)                            

      Miller indices:- of a plane are a set of 3 integers used to designated crystal plane . they are reciprocals of the intercepts of the plane with the axes of the lattice, expressed in their smallest whole number ratios.

      To find the indices of a plane, we find its intercepts with the axes along the basis vectors a, b & c. Let these intercepts be x, y & z. Usually x is a fractional multiple of a, y is a fractional multiple of b and so forth. This forms the fractional triplet .

      Invert this to obtain . and then reduce this set to a similar one having the smallest integers by multiplying it with the LCM of the denominators. This last set is called the Miller Indices of the plane and is indicated by (hkl)

      Expression for Inter planar spacing of cubic crystal

      Consider a plane ABC, in a space lattice with lattice constants a, b & c with intercepts.

     

      If (hkl) is the miller indices of the plane and n is the integer multiplier used, to make the ratio a whole number, we can write

      h=

      k=      ⋯⋯⋯⋯⋯ (2)

      l=

Substituting (2) in (1)

OA=n

OB=n     ⋯⋯⋯⋯⋯⋯(3)   

      OC=n

      Let d_hkl be the normal distance from the origin to the (hkl) plane and ∝,β &γ be the angle made by the normal with the axes. Then

From right triangle ONA: d_(hkl)= OA cos∝

                      From  ONB: d_(hkl)= OB cos β    ⋯⋯⋯(4)

                   & from ONC: d_(hkl)=OC cos γ      

      For an orthogonal lattice,

      Cos²∝+ cos² β+cos² γ=1 ⋯⋯⋯⋯ (5)

      From (4) and (5)

      Using (3),

      For a cubic lattice where a= b = c, eqn. (6) becomes,

      For adjacent lattice planes, n=1

7. Distinguish between Fresnel and Fraunhoffer diffraction. (KU 2006,2005, CU 2009)

Fresnel diffraction	Fraunhoffer diffraction
The source of light or the screen both are at finite distance from the obstacle causing diffraction	The source of light & the screen are at infinite distance w.r.to the obstacle causing diffraction
The wave front meeting the obstacle is spherical or cylindrical	The wave front meeting the obstacle is plane
Lenses are not used	Two convex lenses are used
Eg. Diffraction at a straight edge by a line source	Eg. Diffraction of light through a plane transmission grating

8. Define unit cell and space lattice?                (KU MAY 2005)

      Unit Cell: the smallest unit of a crystal which on repeated addition results in the crystal. Its volume is an integral multiple of volume of primitive cell.

      Space lattice: It is mathematical abstraction used to represent the periodically with which the matter is disturbed in the crystal. Cells which have the smallest area and only one lattice points are known as primitive unit cell

9. What are Newton’s rings?                            (KU MAY 2005)

      When a Plano- convex lens with its convex surface is placed on a plane glass plate, an air film of gradually increasing thickness is formed between the two. These alternate bright and dark fringes are called Newton’s rings.

10. Explain the phenomenon of interference of light? What are the necessary conditions to get clear and distinct    interference fringes?          (KU MAY 2005)

The re-modification of light energy due to the superposition of two light waves of the same amplitude, same frequency and of constant phase difference is called interference. At the points the region where the two light wave arrive in the same phase, the resultant intensity is maximum and the interference is construction. At points where the two light waves arrive in phase difference (opposition), the resultant intensity is minimum and the interference is said to be destructive. It is discovered by Thomas young.

Conditions of interference

There should be two coherent sources of light emitting light waves of same frequency and same amplitude with a constant phase difference.

The light waves from the coherent sources should super impose at the same times and at the same place.

The two coherent sources of light should be very close to each other.

The source must be coherent should have same wavelength, same frequency, constant phase relationship with each other.

11. Define Resolving power of optical instrument. Derive an expression for resolving power of the plane diffraction grating. (KU MAY 2005)

      The ability of an optical instrument to form distinctly separate images of two objects very close to each other is called is resolving power. The resolving power of grating is its ability to show two neighboring spectral lines in a spectrum as separate.

If λ and λ+dλ are wavelengths of two neighboring spectral lines, the resolving power of the grating is defined as the ratio. Suppose there are total N₁ slits in a grating. Let the path difference between the waves from the extreme slits change by λ when the reach any point on the screen. Then the path difference between the waves from the adjacent slits changes by λ/N₁

Considering the interference of waves from adjacent slits the condition for the nth order principal maximum for λ+dλ is (a+b) sinθ = n(λ+dλ)

(a+b) is grating element & θ is the angle of diffraction. The condition for the first minimum of nth order for λ is

(a+b) sinθ = nλ + λ/N₁

Comparing 1 and 2

n(λ+dλ) = nλ+λ/N₁

λ/dλ = nN₁

      the resolving power is proportional to the order of the spectrum n and the total number of lines N_1.

12. How can Newton rings experiment be used to determine refractive index of liquid?

(CU 2010)

 The plano-convex lens is placed on an optically plane glass plate so that the air film is formed between the curved surface and glass plate. Monochromatic light from a sodium vapour lamp is allowed to fall normally on this with the help of a glass plate G inclined at 45⁰. The interference take place and as a result large number of alternate dark and bright rings are obtained.

For air film D²_n+k-D²_n = 4KRλ            à1

            R- Radius of curvature of the curved surface

             λ – Wavelength of light used

Now the plano- convex lens is raised, few drops of given liquid are dropped and the lens is again placed so that a liquid film is formed. Newton’s rings are observed and diameter of each ring is measured.

For liquid film                    d² _n+k –d²_n = 4KRλ/μ à 2

From equation 1 & 2                        μ = D²_n+k-D²_n/d² _n+k –d²_n

The refractive index of the liquid can be measured using this formula.

13. What are the differences between unit cell & primitive cell? (CU 2010)

Unit cell	Primitive cell
Unit cell is the smallest fundamental building block with all the characteristics & the repetition of which forms the entire crystal.	It is the smallest building block, the repetition of which in three dimensions fills the entire space and it is equivalent to one lattice point.
It is the basic structure unit.	It has the smallest volume containing one lattice point.
The lattice is made up of the repetition of unit cell.	It is also unit cell which contain one lattice point only at the corners.
In three dimension a unit cell may be a cube formed by the fundamental translation vectors a,b,c  .	The axial lengths a,b,c will be minimum and they are called primitives.
All unit cells are primitive cells.	All primitive may be or not unit cells.

14. Write the Bragg diffraction equation in reciprocal lattice?

Consider a nano chromatic beam of X-rays of wavelength content on an atom or along with a glaning angle. It is reflected along with the same glaning angle. Similarly is another ray parallel to incident on or at the same glaning angle. It is reflected along. The distance between d. these reflected rays rein for with each other producing intense beam is drawn perpendicular to drawn perpendicular to

            The equation of Bragg’s law is 2dsinθ =nλ

SQ₁= Q₁T= d Sinθ

The path difference between P,Q,R & PQR = SQ₁+Q₁T

The path difference = 2d Sinθ

The condition for reinforcement of these reflected waves is given by the path difference = nλ  where n is an integer 0,1,2,3 etc.

      2d sin θ = nλ

The equation is called Bragg’s law.

d spacing between two parallel lines in the crystal containing atoms

θ diffraction angle.

n Order of the spectrum

λ wavelength of the spectrum.

15 a. Derive the expression for the distance b/n the atomic plane such as (100) and (110) planes.                                          b.   X rays of wavelength 0.71 A⁰ are reflected from (110) plane of a rock salt crystal (a=2.82 A⁰) calculate the gleaning angle corresponding to second order reflection?

OX, OY and OZ are three rectangular axes with the origin O. consider a reference plane passing through O.

Consider a parallel plane passing through A,B and C as in figure. The intercepts are a/b, b/k and c/l respectively. ON is the normal between this plane & reference plane the normal distance these two planes is the spung between the #D lattice planes and it is inter planer spacing ‘d’. This can be represented in terms of the primetime vectors a, b and c of unit cell. Let the normal makes angles with the crystal axes.

ÐNOX = α, ÐNOY =β and ÐNOZ =γ    cosα = d/(a/h), cosβ=d/(b/k) & cosγ  =d/(c/l)

Cos²α + cos²β+ cos²γ =1

( )²+( )²+( )²=1

d²[(h²/a²)+(k²/b²)+(l²/c²)]=1

d=

for a simple vibic lattice a=b=c, the inter planer separation, d_hkl=

in sc lattice 100 plane cuts X axis and 1/l to Y and Z axes 110 planes act obligively across X and Y axes and 11l to Z

d100=  = a

d110 = =

The separation between 100 and 110 is a and a/

b.  λ=0.71A⁰

     α = 2.82A⁰

    d = = 2A⁰

n=2

2d sinθ =n λ

θ= sin^-1( ) = sin^-1 ( )

θ = 22⁰1

16. A soap film of & 4/3 & of thickness 1.5 x 10^-4 cm is illuminated by white light incident at an angle of 600. The light reflected by it is examined by spectroscope in which it is found a dark band corresponding to wavelength of 5x10^-5 cm. calculate the order of interference of the dark band?

A= 4/3

T= 1.5 x 10^-4 cm

I= 60⁰

λ = 5 x 10^-5

2 at cos r = nλ à

(sin i/sinr )= 4/3;

Sinr= (3sini/4) = 3sin60/4 = 0.6495

R= 40.50⁰

Cos r =0.7604

n = (2atcosr/ )= (2x4 x 1.5 x10^-4 x cos40.50)/ (3 x 5 x 10^-5)

n= 6

18. Explain the resolving power of the plane diffraction grating?

AB represents a slit and BC represents a line. Let ‘a’ be the width of each slit and b the width of each line. Then the distance (a+b) is called the grating element. Points separated by an integral multiple of (a+b) are called corresponding points.

Let a plane wave front be incident normally on the grating. Each part of the wave front passing through the slits sends out secondary waves in all directions. Most of the waves travel straight in the same direction of incident light when focused using a convex lens. This is called the central maximum, the position of the central maximum is the same for all wave length. The central maximum will have the same colour as the incident light.

Let λ be the wavelength and θ be the angle of diffraction with the normal to the grating. They travel along Am and CN. Draw ck Lr to Am. There is no path difference between the waves beyond Ck. Then the path difference between the two waves is Ak.

19. What is an air wedge? Explain how it is used for testing the optical plainness of surfaces?

(CU 2008)

    Two plane glass plates AB and AC are placed such that they are in contact at one end & separated by a small distance at the other end. A wedge shaped air film is formed between them. A part of light is reflected from the top surface of the air film & another part of light is reflected from the top surface of the lower glass plate. These two reflected beams interfere. A system of equielistant, dark and bright bands are observed. The angle between the glass plates is called the angle of the wedge.

Let the nth dark band be formed at D where the thickness of the air film is t₁ & the (n+1)^th dark band be formed at F; where the thickness of the air film is t₂.

Let AD=l₁, AF=l₂.

Tanθ =t₁/l₁= =                                                                      à (1)

θ is small then tanθ ~ θ in radians.

θ = (t2-t1)/(l2-l1)          l2-l1 = β  band with    θ = (t2-t1)/β                         à (2)

Condition for nth dark band

2μtcosr = nλ                                                                                   à (3)

For air μ=1, at normal incidence cost =1                2t1=nλ                         à (4)

For the (n+1)th dark band to be formed at F                      2t2 = (n+1)λ     à (5)

(5)-(4) à 2(t2-t1) =λ                                  t2-t1 = λ/2                                à (6)

Substitute (6) in equation (2)

θ = λ/2β

If the medium between the glass plates has refractive index   θ =

The bands are straight & parallel to the axis of wedge optical flatness of the surface.

Optical flatness of the surface

If the surface are optical plane, the finger are straight and of equal thickness. They are of equal thickness because each fringe is the locus of all points where the thickness of air film is constant value.

Given surface AB to be tested is placed on an optically plane surface AC so that an air wedge is formed enclosing a very thin air film. A beam of monochromatic light is allowed to fall normally on this air wedge. Interference take place & as a result large no of dark and bright bands are obtained

2μtcosr = (2n+1) λ/2

Using this method we can test the flatness of surface up to ^th of wavelength of light used.

20.The Bragg angle corresponding to the first order of reflection from (1, 1, 1) plane in a cubic crystal is 30⁰ when x rays of wavelength 0.175 nm are used. Calculate the lattice constant.

                                                                                                                                    (KU 2011)                                                                                           Ans: θ=30⁰

     n=1

     =0.175×10^-10m

      (h k 1)=(1,1,1)

     2dsinθ=n

   

∴d=0.175×10^-10 × =0.303×10^-10m

  Lattice constant, d=3.03×10^-11m