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Determination of refractive index of a liquid by Newton's rings

2019-12-25T09:15:00.000+05:30

Theory of Newton's rings:

7)Determination of refractive index of a liquid by Newton's rings:

First Newton's rings are ontaobta when there is an air film between the glass plate and the plano-convex lens. The system is placed in a metal container. Measure the diameter of nth and (n+p)the dark ring with the help of traveling microscope.

For air film:

Dₙ² = 4nλR

D²ₙ₊ₚ = 4(n+p)λR

.•. D²ₙ₊ₚ - Dₙ² = 4pλR .....(1)

Now pour the liquid whose refractive index is to be calculated in the container without disturbing the arrangement. The air film is now replaced by liquid. Now measure the diameter of nth and (n + p)th dark ring with the help of traveling microscope.

For liquid film :

Let μ is the R.I. of the liquid

D'ₙ² = 4nλR
μ

D'²ₙ₊ₚ = 4( n+p)λ.R
μ

.•. D'²ₙ₊ₚ - D'ₙ² = 4pλR ......(2)
μ

Eq. (1) / eq. (2) gives

μ = D²ₙ₊ₚ - Dₙ²
D'²ₙ₊ₚ - D'ₙ² .....(3)

Eq. (3) gives the refractive index of the liquid by knowing the diameters of nth and ( n+p )th dark rings with and without liquid.

Newton's rings by with light (coloured rings)

2019-12-25T08:41:00.000+05:30

Theory of Newton's rings:

5)Newton's rings by with light (coloured rings):

Diameter of the Newton's ring depends on the wavelength of light used . White light consists of
( seven colours) continuous range of colours. Each colour possesses a particular wavelength. So each colour produces its own system of rings. Only first , it cannot be seen.

At a point, near the point of contact if for a particular wavelength ( say violet colour ), the condition of brightness will be satisfied, then it gives maximum intensity for that colour. At the same point, for other colours, the condition of darkness may be satisfied. Consequently that point will have average intensity due to red colour. The same is true for other colours.

6) Determination of wavelength of monochromatic light using Newton's rings:

Let S be a monochromatic source of light which is at the focus of a lens L₁. So horizontal parallel rays fall on the glass plate B which is placed at 45° to the horizontal line. The glass plate B, partially reflects these parallel rays normally on the lens L placed on the glass plate G. Interference occurs between the rays reflected from the top and the bottom surfaces of the air film. Newton's rings formed can be viewed with the help of traveling microscope focussed on the air film. The position of traveling microscope is adjusted to get the centre of Newton's rings at the position of the intersection of the cross-wire of the eye piece. The microscope is moved until one of the cross wire becomes tangential to the 20th dark ring . This microscope reading is noted. Then the microscope is moved such that the cross wire becomes tangential to 15th, 10th, 5th dark rings respectively. Reading of microscope for each case should be noted . Microscope reading corresponding to the same rings on the other side of the centre should be noted. The reading are noted in the observation table:

Dₙ = Diameter of nth bright ring

Dₙ² =2(2n-1)λR .....(1)

Dₙ₊ₚ= Diameter of (n+p)the bright ring

Dₙ₊ₚ²= 2[2(n+p)-1]λR .....(2)

Dₙ₊ₚ² - Dₙ² = 4.pλ.R

λ = Dₙ₊ₚ² - Dₙ²

4pR .....(3)

The radius of curvature of lens L can be found out by any simple method. Substituting the value of R and the mean value of P in equation (3), λ can be calculated.

Theory of Newton's rings

2019-12-22T09:34:00.001+05:30

Theory of Newton's rings:-

3)Newton's rings with bright centre ( due to reflected light ).

If the plano-convex lens and glass plate are of different materials of refractive indices μ₁ and μ₂ and a liquid of refractive index μ
( lying between μ₁ and μ₂ ) is introduced between them. Then both the reflections from the film would be similar. So the centre of Newton's rings will be bright. This was shown by scientist young, using oil between crown and flint glass,
( μflint >μoil>μcrown ). If μ₁>μ>μ₂, then reflection from top and bottom surface of the liquid film takes place under similar conditions. In both the cases reflection takes place from the surface of optically rarer medium I (path difference=2t only )
Since at a point t=0; the condition of path difference=nλ =0 satisfies. So , central spot appears bright. If μ₁<μ<μ₂ then the central spot will also be bright. Here, also reflections from top and bottom surface of the liquid film takes place under similar conditions. In both the cases reflection takes place from the surface of optically denser medium. So additional phase change of π radius should be considered at interfering rays at the point of contact satisfies the condition for brightness. So the central spot appears bright.

4) Newton's rings by transmitted light

Newton's rings observed by transmitted light are exactly opposite to those observed by reflected light. In this transmitted light, the central spot is bright . By transmitted light:

The diameter of the dark ring,
Dₙ = √2(2n-1)λR

And the diameter for the bright ring,
Dₙ= 2√nλR

Theory of Newton's rings

2019-12-20T09:18:00.001+05:30

Theory of Newton's rings:

2) Expression for the diameter of Newton's rings by reflected light:

Let a plano-convex lens be placed on a plane glass plate. Let R is the radius of curvature of the lens and C is the centre of curvature. The thickness of the air film at point B is MO(=t) at a distance of ON=rₙ from the point of contact O.

For interference due to reflected light,

The path difference between
two interfering rays = 2μtcos( r+ϴ ) +𝞴 /2

Here, μ= 1 (for air film)
r= 0 (for normal incidence)

Geometrical Optics and Interference
And ϴ = 0 ( for large value of R )

.•. path difference=2t + 𝞴 /2 ...(1)

From figure A) ,
R² = rₙ²+(R-t)²
rₙ²= R²-(R-t)²
rₙ²= 2Rt-t²

Since R >>t, t² becomes very small, so it can be neglected.

rₙ² = 2R.t
2t = rₙ²/R

Putting this value of 2t in eq.(1) we get,

Path difference= rₙ²/R+𝞴 /2 ....(2)

Where, rₙ = radius of the nth Newton's ring.

For dark rings:

The condition of dark ring is

Path difference = (2t+1)𝞴 /2

.•. rₙ²/R+𝞴 /2= (2n+1)𝞴 /2

.•. rₙ²/R=n𝞴

If Dₙ is the diameter of the nth Newton's rings then,

rₙ=Dₙ/2 ⇒ rₙ²=Dₙ²/4

.•. Dₙ²/4R =n𝞴 i.e. Dₙ²=4Rn𝞴

Dₙ = 2⎷nR𝞴

.•. Diameter of dark ring is the proportional to the square root of natural number
( Dₙ ∝ ⎷n)

For bright ring:

For condition for bright ring is that, path difference=n𝞴

.•. rₙ²/R + 𝞴/2=n𝞴

rₙ²/R= (2n-1)𝞴/2 where n= 1,2,3,......

Putting rₙ² = Dₙ²/4

.•. Dₙ²/4R = (2n-1) 𝞴/2

Dₙ² = 4R(2n-1) 𝞴/2

=2𝞴R(2n-1)

Dₙ = ⎷2(2n-1)𝞴R

Diameter of the bright ring is proportional to the square root of odd natural numbers.

Dₙ ∝ (2n-1)

Newton's rings

2019-12-18T07:33:00.000+05:30

Newton's rings:

A plano-convex lens of large radius of curvature is placed with its convex surface on a glass plate. The air film is formed between the lower surface of the lens. AOB and upper surface of the glass plate G. The thickness of the film is zero at the point of contact O and gradually increases outward. If monochromatic light is allowed to fall normally on this film, and the film is viewed for reflected light then alternate bright and dark concentric rings are seen, with their centre as dark . These rings were first studied by Newton and hence are known as Newton's rings. Newton's rings are formed as a result of interference between the light waves reflected from the upper and lower surface of the air film.

1) Formation of Newton's rings:

Newton's rings are formed as a result of interference between the waves reflected from the top and bottom surfaces of the air film.

When a ray of monochromatic light is incident normally on the air film, on reflection from the top and bottom surfaces of the film, the reflected rays interfere. The path difference between them ( ray 1 and 2) will be

2𝜇.t.cos(r+ϴ ) + λ/2

Where 𝜇- Refractive index of the film.
t - thickness at a point under consideration.
r - angle of refraction at the upper face of the air.
ϴ- angle of wedge.

Now, for air film, 𝜇= 1; and for normal incidence r =0. And the large radius of curvature of plano-convex lens;ϴ becomes 0(nearly).

.•. path difference between rays 1&2=2t+ λ/2
...(1)

At the point of contact of the lens and glass plate, t = 0.

.•. from equation (1) path difference = λ/2

This is the condition for minimum intensity. Hence the centre spot of Newton's rings is dark

The condition for maximum intensity is path difference = nλ.

.•. from equation (1) 2t + λ/2 = nλ.

So for nth order, maximum intensity occur for a constant value of thickness t. In air film, thickness ‘t' remain constant along a circle.so brightness is in the form of circle. Different maximum intensity will occur for different values of ‘t'
Similarly minimum intensity will also in the form of circular rings.
The maximum intensity circular rings occur when path difference is λ,2λ,3λ,....

The minimum intensity circular rings occur when path difference is λ/2, 3λ/2, 5λ/2,... i.e. circular rings of maximum and minimum intensities will occur alternatively. Each fringe ( or ring ) is a locus of constant film thickness and so these are the circular fringes of constant thickness.

Thin films

2019-12-15T10:04:00.002+05:30

Thin films:-

Thin films means a transparent layer of very small thickness in the range of 0.5 μm to 10 μm ; such as soap film or a layer of skin oil floating on water surface. When light is imcidein on a thin film, different colours are seen. This can be explained by considering interference between light reflected from the top and the bottom surface of a thin film. In case of thin films, interference occur due to i) reflected light and ii) transmitted light.

Interference in thin film of uniform thickness due to reflected light.

Consider a thin transparent film of uniform thickness ‘t' and of refractive index ‘μ' be placed in air. Let a beam of light AB is incident on the upper surface l, at an angle' i as shown in figure A. At B, light suffers partial reflection along BE and partial refraction along BC. The angle of refraction is r. At C, it is again partially reflected along CD. At D it is partially refracted along DF.

Incident beam AB suffers partial refraction at B ( top surface ) and at C
( bottom surface). So we get a set of parallel beams like BE , DE. etc. These beams originate from same incident beam. So these beams are coherent. When these reflected beams BE and DF superimpose, interference takes place.

Let us calculate the path difference between the beams BE and BCDF. Let DM and BN be perpendicular to BE and CD respectively. Produce BH and DC to meet at G. Let t= BH is the thickness of the film.

By geometry ∠HGC = r ; ∠BDM =i ; ∠DBN= r
.•. BC = GC

The path difference between two reflected beams BE and DF=(BC+CD) in film-BM in air

= μ ( BC + CD ) - BM
= μ ( BC + CN + ND ) - BM ....(1)

We know that,

μ = sin i = BM / BD = BM
sin r ND / BD ND

.•. BM = μ.ND

.•. Equation. (1) becomes, path difference

= μ ( BC + CN + ND ) - μ. ND
= μ ( BC + CN )
= μ ( GC + CN )
= μ . GN

In ∆ BGN,

cos r = GN = GN ( •.• BG = BH+HG = t + t = 2t)
BG 2t

.•. GN = 2.t cos r .....(3)

Putting this value of GN in eq. (2) we get ,

Path difference =2.μ.t.cos r .....(4)

Eq. (4) does not represent a correct path difference. Because whenever reflection takes place from the surface of optically denser medium a phase change of π radius ( or path difference λ/2) takes place.

The film is optically denser than the surrounding medium ( air ).
The ray BE originating by reflection at the denser medium, suffers a phase change of π radius due to reflection at B.

( No such change of phase occur for ray DF, as it is result of reflection at C, the surface of rarer medium ).

The correct path difference between reflected beams = 2μ.t.cos r + λ/2 = nλ

i) The firm appears bright if the path difference. = nλ

2μ.t.cos r + λ/2 = nλ

.•. 2μ.t.cos r = (2n - 1 ) λ/2

ii) The firm appears dark if path difference
= (2n + 1) λ/2

.•. 2μ.t.cos r + λ/2 = (2n + 1) λ/2

2μ.t.cos r = nλ where, n = 0,1,2,....

Interference of light

2019-12-12T22:30:00.002+05:30

Interference of light:-

Interference is the phenomenon exhibited by light when two or more light waves superimpose on each other.

The resultant light intensity at any point in the region of superposition is different from the intensities due to each wave. It means that the resultant intensity of light changes from point to point in the region of superposition. The modification of light intensity in the region of superposition is called interference of light.

In the region of superposition there are some points where the waves superimpose in such a way that the resultant intensity is greater than the sum of the intensities due to individual waves . This is constructive interference. There are some other points where the resultant intensity is less than the sum of the intensities due to individual waves. This is destructive interference.

a) condition for maximum and minimum intensity:

If two light waves having same speed the phase difference between them. Constructive interference is said to take place only when the resultant light intensity is maximum. These points appear to be bright.

When phase difference 𝜹= 0, 2π, 4π, ...., 2nπ

i.e corresponding path difference = 0, λ,2λ,..,nλ.

So the condition for constructive interference is that the path difference between the two waves should be an integral multiple of the wavelength λ.

I.e. Path difference = nπ. Where n = 0,1,2,....

Destructive interference is said to take place only when the resultant light intensity is minimum. So these points appear to be dark.

When phase difference,
𝜹= π, 3π, 5π, ...,(2n+1)π

i.e Corresponding path difference,
= λ,3λ,5λ,...,(2n+1) λ
2 2 2 2

If the amplitudes of the waves are not equal, there will not be complete destructive interference. The condition for destructive interference is that the path difference between the two waves should be an odd integral multiple of half the wavelength.

i.e. path difference= (2n+1) λ
where n=0,1,2,3,....

Thus as the position of the point P on the screen changes, the path difference between the two waves at P changes and the intensity becomes alternatively maximum and minimum. This is called interference pattern.

b) Classification of interference phenomenon:

Broadly interference phenomenon can be classified into two groups.

The important condition for steady interference pattern is that the two light source must be coherent. It means the phase difference between them is always constant. Generally such type of light source can be obtained from a single and the same source of light by using different methods. There are two methods for of light by using different methods. There are two methods for producing such coherent sources. These are,

i) Division of wavefront and
ii) Division of amplitude.

Depending on a particular method, the phenomenon of interference is studied. And so interference phenomenon can be studied under these two categories. i) Division of wavefront method and ii) Division of amplitude method.

i) Division of wavefront method :

In this method, the wavefront generated by a particular source ( known as a incident wavefront ) is divided into two separate wavefronts by using some techniques as in case of superimpose on each other gives a stedy interference pattern. Young's experiment , Fresnel's biprism experiment are the examples of interference by division of wavefront.

ii) Division of amplitude method :

In this method, the amplitude of incident wave is divided into two beams by using partial reflection and refraction. ( As in case of Michelson's experiment). Then by using some techniques these two beams superimpose on each other gives a stedy interference pattern. Michelson's interferometer, interference in thin films, Newton's ring experiment are the examples of interference by division of amplitudes.

Equivalent focal length of coaxial lens system

2019-12-11T12:29:00.001+05:30

Equivalent focal length of coaxial lens system:

Consider two thin lenses L₁ and L₂ focal length f₁ and f₂ respectively. These are placed in air separated by a distance ‘a'. Let a light ray AB is incident on lens L₁, parallel to the principal axis at a height X above the axis. After refraction of this ray through lens L₁, it deviates through an angle δ₁ and the emergent ray ( in the absence of L₂ ), passes through second principal focus of L₁,
Which is at D. The ray AB is deviated through an angle δ₁ by the lens L₁. In figure A.

Let, O₁B = x , O₁O₂ = a, O₂ C = y

F₁ and F₂ are focal points of the optical system.
P₁ and P₂ are principal points of the optical system.
δ₁ = tan δ₁ = O₁B = X
O₁D f₁

.•. Derivation produced by first lens = δ₁= x
f₁

The emergent ray from the lens L₁ meets the lens L₂ at C which is at a height y above the axis. After refraction through lens L₂ the emergent ray meets the axis at F₂ which is second focal point of the lens system. The lens L₂deviated the ray BC further through δ₂.

Geometrical Optics and Interference

.•. Deviation produced by the second lens
= δ₂ = y / f₂

When AB , the incident ray and CF₂, the emergent ray are produced, they meet at E. Then a single convex lens placed at EP₂ having focal length P₂F₂ ( = f ) is equivalent focal length of the lens system.

.•. Deviation produced by equivalent lens
= δ = EP₂ / P₂F₂

δ = x / f

Deviation produced by equivalent lens system is the algebraic sum of deviation produced by individual lens.

.•. δ = δ₁+ δ₂

.•. x = x +  y
f f₁ f₂ ......( 1 )

From figure A.

y = HO₂ - HC
= x - a. δ₁
{since angle = arc / radius
δ₁ = HC/ BH = HC / a }
= x - a. x /f₁

y = x (1- a/f₁ ) .....( 2 )

Putting this value of y in equation (1)

  x = x +  y (1- a  )
f f₁ f₂ f₁

   1= 1 +  1 (1- a  )
f f₁ f₂ f₁

  1= 1 +  1  -    a     ..... (3)
f f₁ f₂ f₁f₂

Where f is the equivalent focal length of coaxial lens system.

a) power of lens :

Power of a lens is its ability to converge or diverge the rays incident on it. The power of lens is the reciprocal of its focal length in meters.

Power of lens = 1
Focal length in meters

The unit of power of lens is diopter.

A convex lens of small focal length produces a large converging effect while a convex lens of large focal length produces a small converging effect. So power of a convex lens is taken as positive. A concave lens produces divergence. Therefore, power of cancave lens produces divergence. Therefore , power of concave lens is taken as negative.

b) power of coaxial lens system:

When two thin lenses of focal lengths f₁ and f₂ are placed coaxially and separated by a distance "a" , then equivalent focal length f of lens system is given by

  1= 1 +  1  -    a
f f₁ f₂ f₁f₂

But, power =    1
Focal length( in meters )

.•. P = P₁ + P₂ - aP₁P₂

.•. where P is the equivalent power of coaxial lens system.

c) Position of cardinal point :

(i) Second principal point

Let β = O₂P₂ = Distance of the second principal point from the second lens.

And ɑ = Distance of the first principal point from the first lens.

The calculate the value of β we have to produced as follows: from figure .A.

O₂P₂ = P₂F₂ - O₂F₂ = f - O₂F₂ .....(4)

Now , the triangles, ∆EP₂F₂ and ∆CO₂F₂ are similar Geometrical Optics and IntererencIn

.•. EP₂ = P₂F₂ i.e x = f .
CO₂ O₂F₂ y O₂F₂

  O₂F₂ = y. f
x

.•. putting the value of ‘y' from equation (2) we get ,

  O₂F₂ = x (1- a  )f = ( 1- a  )f
f₁ f₁
x

putting the value of O₂F₂ in equation ..(4) we get,

O₂F₂ = f - (1-a  )f = a . f
f₁ f₁

Since P₂ lies on the left side of lens L₂.

•.• O₂F₂ = - β

.•. β = - a. f
f₁ .....(5)

So second principal point ( plane ) P₂ is at a distance "β " to the left from the second lens.

ii) First principal point : To calculate the value of ɑ we have to consider rays parallel to principal axis from the right hand side as shown in figure B.

Proceedings in the similar manner as described above we get,

O₁P₁ = ɑ = a.f
f₂ .....(6)

So, first principal point P₁ is at a distance "ɑ" to the right from the first lens.

Now, as the medium on the two side of the lens system is same, the nodal points N₁ and N₂ coincide with P₁ and P₂.

iii ) Second focal point : The distance of the second focal point F₂ from the second lens L₂ will be O₂F₂ and is given by.

O₂F₂ = P₂F₂ - P₂O₂
= f - (- O₂P₂ )
= f + O₂P₂
= f + β
= f + (- af )
f₁
= f + ( 1- af )
f₁

iv) First focal point : The distance of the first focal point F₁ from the first lens L₁ will be O₁F₁ and is given by

O₁F₁ = P₁F₁ - P₁O₁
= -f - (-O₁P₁ )
= -f + O₁P₁
= -f + ɑ
= - f + af
f₂
= - f ( 1- a)
f₂

Geometrical Optics and Interference

Let O = Positive of object
I = Positive of image
P₁ = First principal point
P₂ = Second principal point
u = O₁O= Object distance from the first lens.
v =O₂I= Image distance from the second lens.
U= P₁O= Object distance from the first principal point.
V= P₂I= Image distance from the second principal point.
ɑ= Distance of the first principal point from first lens.
β= Distance of the second principal point from second lens.

From figure, ( using sign convention)

-U = -u +ɑ
U = u - ɑ
And. V= v -β

For lens combination

1 _ 1 = 1 and magnification m = V
V U f U

Cardinal points of an optical system

2019-12-09T22:17:00.001+05:30

Cardinal points of an optical system:

A combination of lenses having common principal axis is called a coaxial system of lens. In such systems, a method for finding out the distance and size of the image is very tedious. Gauss showed that a coaxial system can be considered as one system. For any system of lenses, there are six points known as cardinal points.

They are:

i) two focal points F₁ and F₂
ii) two principal points P₁ and P₂
III) two nodal points N₁ and N₂.

Consider a coaxial optical system of two convex lenses of focal lengths f₁ and f₂ placed at a distance ‘a' apart as shown in figure.

Focal points :

To explain the first focal point on the object side refer figure. A. Consider an incident ray of light AB passing through axial point F₁. After refraction through lens system this ray becomes parallel to the principal axis ( in a direction CD ). Then this axial point F₁ through which rays are incident on the lens system is known as first focal point. A plane passing through F₁ and perpendicular to the principal axis is known as first focal plane.

Focal point

To explain second focal point on image side refer figure.B. consider incident rays parallel to the principal axis. After refraction through the lense system, these rays focus on the axial point F₂, then this axial point F₂ is known as second focal point. A plane passing through F₂ and perpendicular to the principal axis is known as second focal plane.

Principal points:

In figure .A. incident ray AB produced forward and emergent ray CD produced backward to meet at E. A vertical plane passing through this point E and perpendicular to the principal axis is known as first principal plane. The point where the first principal plane intersect the axis is known as first principal point. Here P₁ is the first principal point. The distance P₁F₁ between first principal point and first focal point is known as the first principal focal length of the system.

In figure B. Incident ray GH produced forward and emergent ray KJ produced backward to meet at M. A vertical plane passing through this point M and perpendicular to the principal axis is known as second principal plane. The point P₂ where the second principal plane intersect the is known as principal point.

Principal points are the points having unit positive magnification for lens system. The distance between first focal point and the first principal point is known as first principal focal length ( P₁F₁ ) of lens system. Similarly the distance between second focal point and second principal point is known as a second principal focal length ( P₂F₂ ) of lens system.

Nodal points :

Nodal points are the two conjugate points on the axis of lens such that the relative angular magnification is unity. When the light ray is incident at an angle Rita , on the first nodal point N₁, emerges from the optical system at the same angle Rita from the another nodal point N₂. It means these are the points on the axis having unit positive angular magnification. So, the nodal points have a property that the rays , through these points are parallel to each other. The plane which passes through nodal points and perpendicular to the axis XX' are known as nodal planes.

When the medium on both sides of optical system in same, then the nodal points and principal points coincide.

Gravitational potential due to a uniform solid sphere

2019-12-06T19:21:00.001+05:30

Gravitational potential due to a uniform solid sphere:

1) At a point outside the sphere:

Consider a uniform solid sphere having radius R and mass M. Let P be a point at a distance ‘r' from the centre Of the solid sphere, where the gravitational potential due to the sphere is to be determined.

Imagine the solid sphere to be made up of a large number of thin concentric spherical shell's of masses m₁, m₂, m₃, ..... etc. Now, the gravitational potential at P due to whole sphere is equal to sum of potentials due to all such shells.

.•. V = - [ Gm₁+ Gm+ .....]
r r

= - G ( m₁+ m₂ + .....)
r

.•. V = - GM
r

Where, m₁+m₂ + .... = M = Mass of the sphere.

Intensity of gravitational field due to solid sphere:

E = - dV = - d ( - GM )
dr r r

.•. E = - GM
r²

Negative sign shows that, the intensity of gravitational field is directed towards the centre O of the sphere.

2) At a point on the surface of the sphere:

At a point on the surface of the sphere, r=R :

V = - GM and
R
E = - GM
R²

3) At a point inside the sphere :

Consider a uniform point P inside the sphere at a distance‘r' from its centre O , as shown in figure.

Let R = radius of sphere and
ϱ = density of the sphere

Now, with O as a centre and radius OP = r drawn a sphere.

Then the point P lies on the surface of the solid sphere of the radius ‘r,' and inside the spherical shell of internal radius R and external radius R.

.•. volume of inner solid sphere = 4 π r³
3

.•. Mass of inner solid sphere = 4 π r³ϱ
3

The potential V₁ at P due to inner solid sphere is :

V₁ = - G ( 4πr³ ϱ ) = 4πr² ϱ G ..... (1)
r

To find the potential V₂ due to the outer spherical shell, draw two concetric spheres of radii ‘x' and ( x + dx ) , forming a thin spherical shell of thickness ‘ dx '.

.•. Volume of shell = 4πx² dx

.•. Mass of shell = 4πdx²dx

.• Potential at P due to this shell

P = - G (4πdx²dx ) = -4πdGx dx
x
.•. The potential V₂ at P due to thick spherical shell is obtained by integrating above equation between the limits x=r to x=R.

V₂ = -4πϱG ᷊∫  ᷢᷢᷢᷢᷢᷢᷢᷢx dx = -4πϱG [x²/2] ᷊ ᷢᷢᷢᷢᷢᷢᷢᷢ

.•. V₂ = -4πϱG [ R² - r²]
2

.•. V₂ = -2πϱG ( R²-r² ) ......(2)

.•. Total potential at P is given by :

V = V₁+V₂

.•. V = - 4 π r²ϱ G-2πϱ G(R²-r²)
3

.•. V = - 2πϱ G(2 r² + R² - r²)
3

.•. V = 2πϱ G(3R²-r²)
3

.•. V = -G( 4  π R³ϱ ) [3R²-r²/2R³]
3

V = - GM [ 3R³- r²] .......(3) ,
2R³
Since,  4 πR³ϱ = M
3

This is an expression for the potential inside a solid sphere, at a distance, from its centre.

The minimum potential at the centre of the sphere can be obtained by putting r=0.

V₀= -3GM ......(4)
2R

The intensity of gravitational field at P inside the solid sphere is:

E = - dV = -  d [-GM ( 3R²-r²)]
dr dr 2R³

E = - GM [2r]
2R³

.•. E = - GM   ... .... (5)
R³

Gravitational potential due to a spherical shell

2019-12-05T09:12:00.000+05:30

Gravitational potential due to a spherical shell:

1) At a point outside the shell:

Consider a uniform spherical shell of radius R.

Let P be a point outside the spherical shell at a distance ‘r ' from the centre O of the shell.

Let ϱ = Mass per unit area of the surface.

As shown in figure, two planes AD and BC cut the shell vertically.

The element between the two planes is a slice ABCD in the form of a ring of small angular width dϴ with the OP as axis.

Each element of the ring is at a distance , AP = x from the outside point P.

.•. Thickness of the shell, AB = R dϴ

Radius of the shell, CK = R sinϴ

.•. Surface area of the slice= (2πR sinϴ) (R dϴ ) = 2πR² sinϴ dϴ

.•. Mass of the slice= Surface area of slice× Mass per unit area (ϱ)

Figure: Gravitational potential at a point outside the spherical shell.

.•. Mass of the slice, Mring = 2πR²sinϴ dϴ ϱ

.•. potential at P due to the ring is :

dV = - GMring
X

= - G(2πR²ϱsinϴ dϴ )/x .....(1)

In ∆ APO : x² = r² + R² - 2rR cosϴ

Differentiating above equation, we get :

2x dx = 2rR sinϴ dϴ
( since,R and r are constant )

.•. x = rR sinϴ dϴ
dx

Substituting this value of ‘x' in eq. (1) , we get :

dV = - G ( 2πR²ϱsinϴ dϴ ) dx
rR sinϴ dϴ

= - 2πRϱG dx

Integrating above equation for the whole shell between the limits,

x = ( r- R ) to x = ( r+R ) , we get :

But, 4πR²ϱ = M ( Mass of the whole shell )

.•. V = - GM
r ...... (2)

Thus, for a point outside the shell, the shell behabes as if the whole mass is concentrated at the centre of the shell.

Now, the intensity of gravitational field at a point outside the shell is given by:

E = - dV / dr

= - d / dr ( - GM / r )

.•. E = - GM / r² ...... (3)

Negative sign shows that intensity of gravitational field is directed towards the centre O of the shell.

(2) Potential on the surface of the shell :

For a point on the surface of the shell, r= R.

.•. Potential , V = - GM / R

Intensity of the field, E = - GM / R²

Guass’s theorem for gravitation

2019-12-04T08:05:00.001+05:30

Guass’s theorem for gravitation:

Guass's law states that the total gravitational flux over a closed surface, having a unique outward drawn normal to it at every point is - 4πG times the total mass enclosed by that surface.

The gravitational field E at a distance ‘r' from a point mass M is given by :

E= - GM/r² .....(1)

Then the flux (ϕ) of the gravitational field, through the surface of the sphere of radius ‘r' is given by :

ϕ = - GM × 4πr² = - 4πGM

r²

As shown in figure, this flux ϕ is due to the normal component of gravitational field

E = - GM cosϴ

r²

Let , dϕ = small flux through an element of area dA.

.•.dϕ = -GM cosϴ dA

r²

.•. dϕ= E.dA = n. E dA

.•. Total gravitational flux enclosed by the closed surface is :

ϕ = ∫ E. dA = ∫ n^. E dA

.•. ϕ = ∫ E1.dA + ∫ E2.dA + ........

.•. ϕ = -4πG (M1 + M2 + ......)

.•. ϕ = -4πGM

where , M = sum of all masses enclosed by the surface.

This is Gauss's law in gravitation.

Gravitational potential energy

2019-12-03T19:07:00.000+05:30

Gravitational potential energy:

The amount of work done in moving a unit mass from infinity to the point under consideration , in the gravitational field of the body is called gravitational potential at that point.

The gravitational potential is scalar quantity and it's S.I unit is joule/kg.

Gravitational potential due to a point mass :

Consider a point A at a distance ‘r' from a particle of mass M, as shown in figure.

Then intensity of Gravitational field at point A is given by:

E = - GM/r² (along AM )

.•. Difference of Gravitational potential between two points A and B at a distance ‘dr' apart is :

VA-VB= dV = - E dr .......(1)

.•. E = - dV/dr ....... (2)

This is the relation between gravitational field and gravitational potential.

.•. the gravitational field is defined as the negative gradient of gravitational potential.

Now, from eq. (1) , we get:

dV = - ( - GM/r²) dr = GM/r² dr

Hence, potential V at point A ( i.e work done in moving unit mass from infinity to point A) in the gravitational field of M is given by:

Thus, gravitational potential is zero at infinity. And at all other points it is less than zero i.e it has a negative value.

Gravitational potential energy:

The gravitational potential energy of a body at a point in the gravitational field is defined as the amount of work done in bringing the body from infinity to the desired point.

Consider a point A at a distance ‘r' from a body of mass M. A particle of mass ‘m' is placed at B , at a distance (r+dr ) in the gravitational field of M.

Gravitational force of attraction acting on mass ‘m' is :

F = - GMm/r² (along AM )

Hence, the work done (dW) by the force F in moving the mass‘m' from B to A , through a small distance ‘dr' is given by:

dW = - F dr = - (- GM/r²) dr

.•. total work done in moving the mass ‘m' from ∞ to point A is given by:

.•. W = - GMm/r .....(1)

This work done is stored in the body as it's gravitational potential energy (U) and it is given by :

U = - GMm/ r .......(2)

.•. Gravitational P.E (U) =gravitational potential (V) × mass of the body (m)

Gravitational potential energy (U) is always negative . It is a scalar quantity and it's SI unit is joule ( J ).

Gravitational field

2019-12-03T17:06:00.000+05:30

Gravitational Field

The gravitational field is the space around a body within which any other mass experience a gravitational force of attraction.

Intensity of gravitational field at a point:

It is defined as the gravitational force acting on a unit point mass, placed at a given point, in the gravitational field of the body.

Consider a body of mass M which sets up the gravitational field in the space surroundings it.→

Now, a small test mass ‘m' is placed at a point P at a distance ‘r' from the mass M. The test mass ‘m' experience an attractive gravitational force F.

Then the intensity of gravitational field E at point P is given by: E=F/m

The intensity of gravitational field is a vector quantity.

The direction of E is same as that of F.

The magnitude of gravitational field strength at point P is :

.•. E= F/m = - GM/r² .....(1)

since, F= - G Mm/ r²

Negative sign shows that the field is directed towards the mass M , In vector form, the intensity of gravitational field can be given as :

E= GM/r² r^ ....... (2)

Where, r^= unit vector along the line of force.

It's SI unit is N/kg and

dimensions are [M°L¹T-²].

Acceleration due to gravity (g)

2019-12-02T19:24:00.001+05:30

Acceleration due to gravity (g):

Consider the earth as a solid sphere of mass M and radius R. Let a small body of mass ’m' is placed at a height ‘h' from the surface of the earth, as shown in figure. According to Newton's law gravitation, the force exerted on a body by the earth is given by:

F = G Mm ......(1)
(R+h)²

All . freely falling bodies experience a constant acceleration, due to Gravitational pull of the earth. This is called acceleration due to gravity (g). The acceleration that results due to earth's gravity is called acceleration due to gravity (g).

It is directed downwards towards the centre of the earth.The value of ‘g’ varies from a minimum at the equator and the maximum at the poles.

Now, according to Newton's second law of motion:

F=mass×acceleration=m gh .....(2)

Where, gh= acceleration due to gravity at a height ‘h’.

From eq. (1) and (2) , we get:

m gh = G Mm
(R+h)²

.•. gh = GM ......(3)
(R+h)²

On the surface of the earth, h=0 and gh=g.

.•.acceleration due to gravity (g) on the surface of the earth is :

g = GM .....(4).
R²

This is the relation between G and g.

Expression for ‘g’ at a height ‘h’ above the earth's surface:

The Gravitational force exerted by the earth on a body of mass ‘m' lying on its surface is given by:

mg = GMm .......(1)
R²

If the body is at a height ‘h' above the earth's surface, then we have:

m gh = GMm .....(2)
(R+h)²

Where, gh = acceleration due to gravity at a height ‘h' from the earth's surface.

Eq. (2) by eq. (1) gives:

gh = R² = R² = (1+h/R)²
g (R+h)² R²(1+h/R)²

Expanding and neglecting higher powers:

gh/g= (1-2h/R)

.•. gh= (1-2h/R)g .....(3)

Hence, as altitude ‘h' increase the value of ‘g' decrease.

Expression for ‘g' at a depth ‘d' below the earth's surface:

Let p = density of the earth.
M = Mass of the earth having radius
R= 4πR³p/3

.•. The Gravitational attraction on a mass ‘m' lying on earth's surface is:

mg = GMm/R² = G×4/3πR³p×m/R²

= 4/3 ×πRGm ...... (1)

Where , g = acceleration due to gravity on the earth's surface of the earth.

Now, the Gravitational attraction on a mass ‘m' at a depth ‘d' is due to inner spherical part of the earth having radius
(R-d).

mgd= G 4/3 × π(R-d)³ρ×m
(R-d)²

= 4/3π(R-d) Gρm .....(2)

Where , gd = acceleration due to gravity at a depth‘d' below the earth's surface.

Eq. (2) by (1) gives gd/g = (R-d)/ R

.•. gd = (1-d/R)g    .....(3)

Hence , the value of gd goes on decreasing with depth and at the centre‘g' will be zero.

Newton's law of gravitation

2019-12-01T07:22:00.002+05:30

Newton's Law of Gravitation:

Every particle of matter in the universe attracts every other particle with a force which is directly proportion -al to the square of the distance between the two particle.

Hence, the gravitational force of attraction between two point masses ‘M’ and ‘m' separated by a distance ‘r' is given by:

F ∝ Mm and F ∝ 1/r² .•. F ∝ Mm/r²

F=-G Mm ..........(1)
r²

Where, G= Gravitational Constant ,
M and m are two point masses, separated by a distance ‛r’.

The negative sign shows that the Gravitational force is always attractive.

Gravitation Constant G : In eq. (1) if, M=m and r=1 then F=G (Numerically).

Definition: The universal G gravitational constant (G) is equal to the force of attraction between two bodies each of unit mass, lying ar a unit distance apaet.

G=6.67×10-¹¹ N.m²/kg² in S.I units

The S.I Unit of G is newton.m²/kg²
or N.m/kg²

Dimension of G is : [M¹L¹T-²] [L²]
[M²]

= [M-²L³T-²]

Kepler's Laws, Kepler's first law of elliptical orbis, Kepler's second law for areal velocity, Kepler's third law of period

2019-11-30T14:12:00.000+05:30

Kepler's Laws:

Kepler's Laws of planetary motion:

Kepler formulated following three laws concerning the motion of the planets around the sun.

(1) Kepler's first law of elliptical orbits:

All planets move in an elliptical orbits, with the sun at its foucs.

Proof: The eccentricity (ε) of the orbit of a particle , moving under attractive central force is given by:

ε²=1+ 2EJ÷mC² E= - mC²÷2J²(1-ε²)

Where , J = Angular momentum of the particle of reduced mass ‛m’.
E = Total energy of a particle.
C = Constant= GMm
(in case of gravity force).
M= Mass of the body about which the particle moves.

In the case of a planet revolving round the sun , the sun is at one foucs which is takes as the centre of the coordinate system of the ellipse. If ε <1 then the total energy of the system ( planet and the sun ) is negative and the planet remains bound to the sun .

As the planet is bound to the sun and cannot escape from it, the planet moves around the sun , in closed elliptical orbit. This proves Kepler's first law of elliptical orbits.

(2) Kepler's second law for Areal velocity:

The speed of a planet along its orbit varies in such a way that the radius vector drawn from a sun to a planet sweeps out equal areas of the orbit in equal interval of times.

In short, the Areal velocity of the planet around the sun is Constant.

As shown in figure , the planet moves from the position a to b and from the position c to d in the same time. Then two shaded areas swept out by a planet in equal times are equal.

Proof: Suppose a planet of mass ‛m’ is moving in an elliptic orbit around the sun as shown in figure . Let S be the centre of the sun.

Let‛r’ be the radius vector of the planet with respect to sun. Suppose the planet moves from P to P' in a small time dt.

The area swept out by the radius vector is the area SPP'.

If ‛dt’ is infinitesimally small then PP' is a straight line = r dθ

Now , SPP' is a triangle.

.•. Area of ∆ SPP'=dA = 1 r . rdθ = 1 r²dθ
2 2

This is the area swept out in time dt.

.•. Rate at which the area is swept is Called Areal Velocity of the planet.

.•. Areal Velocity , dA = 1 × r² dθ ........(1)
dt 2 dt

Now , the Angular momentum of the planet , J = mvr = mwr² is a constant under a central force .

.•. ω = J/ mr² or dθ/dt = J/mr²

Hence , eq.(1) becomes:

dA = J/2m = constant .........(2)
dt

Thus , the Areal velocity of the planet around the sun is Constant. This proves the Kepler's second law that the radius vector joining the sun to the planet sweeps out equal areas in equal intervals of time.

(3) Kepler's third law of period:

The square of the time period of any planet about the sun is proportional to the cube of the planet's mean distance from the sun i.e semi-major axis of the elliptical orbit of the planet.

The time period(T) of the planet(i.e time taken by it to complete one revolution around the sun is given by:

T= Area of the ellipse
Areal velocity of the radius vector

= πab
J / 2m

Where , a and b are the semi- major and the semi-minor axes of the elliptical orbit of the planet respectively.

.•. T² = 4π²m²a²b²
J² ......(1)

Now, if l is the letus rectum of the ellipse, then we have :

l=b²/a .•. b²= la

Hence, eq. (1) becomes:

T²= 4π²m²a³l
J²
But, 4π²m²l = Constant
J²

.•. T² ∝ a³ ....(2)

Hence, the square of the time period of the planet is proportional to the cube of semi- major axis of the ellipse . This is Kepler's Third Law of Period.

Learn Easy With Science

Determination of refractive index of a liquid by Newton's rings

Theory of Newton's rings:7)Determination of refractive index of a liquid by Newton's rings:

For liquid film :

Newton's rings by with light (coloured rings)

Theory of Newton's rings:

5)Newton's rings by with light (coloured rings):

6) Determination of wavelength of monochromatic light using Newton's rings:

Theory of Newton's rings

Theory of Newton's rings:-

3)Newton's rings with bright centre ( due to reflected light ).

4) Newton's rings by transmitted light

Theory of Newton's rings

Theory of Newton's rings:

2) Expression for the diameter of Newton's rings by reflected light:

Newton's rings

Newton's rings:

1) Formation of Newton's rings:

Thin films

Thin films:-

Interference in thin film of uniform thickness due to reflected light.

Interference of light

Interference of light:-

Equivalent focal length of coaxial lens system

Equivalent focal length of coaxial lens system:

a) power of lens :

c) Position of cardinal point :

Cardinal points of an optical system

Cardinal points of an optical system:

Focal points :

Principal points:

Nodal points :

Gravitational potential due to a uniform solid sphere

Gravitational potential due to a uniform solid sphere:

Consider a uniform solid sphere having radius R and mass M. Let P be a point at a distance ‘r' from the centre Of the solid sphere, where the gravitational potential due to the sphere is to be determined.

Gravitational potential due to a spherical shell

Gravitational potential due to a spherical shell:

(2) Potential on the surface of the shell :

Guass’s theorem for gravitation

Guass’s theorem for gravitation:

Guass's law states that the total gravitational flux over a closed surface, having a unique outward drawn normal to it at every point is - 4πG times the total mass enclosed by that surface.

Gravitational potential energy

Gravitational potential energy:

Gravitational potential due to a point mass :

Gravitational potential energy:

Gravitational field

Gravitational Field

Intensity of gravitational field at a point:

Acceleration due to gravity (g)

Acceleration due to gravity (g):

All . freely falling bodies experience a constant acceleration, due to Gravitational pull of the earth. This is called acceleration due to gravity (g). The acceleration that results due to earth's gravity is called acceleration due to gravity (g).

Expression for ‘g’ at a height ‘h’ above the earth's surface:

Expression for ‘g' at a depth ‘d' below the earth's surface:

Newton's law of gravitation

Newton's Law of Gravitation:

Every particle of matter in the universe attracts every other particle with a force which is directly proportion -al to the square of the distance between the two particle.

Gravitation Constant G : In eq. (1) if, M=m and r=1 then F=G (Numerically).

Definition: The universal G gravitational constant (G) is equal to the force of attraction between two bodies each of unit mass, lying ar a unit distance apaet.

Kepler's Laws, Kepler's first law of elliptical orbis, Kepler's second law for areal velocity, Kepler's third law of period

Kepler's Laws:

(1) Kepler's first law of elliptical orbits:

(2) Kepler's second law for Areal velocity:

(3) Kepler's third law of period:

Theory of Newton's rings:

7)Determination of refractive index of a liquid by Newton's rings: