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IIT JEE Study Material - Electrostatics - Capacitor & Combinations CAPACITORS Capacitance If Q is the charge given to a conductor and V is the potential to which it is raised by this amount of charge, then it is found that Q V. Or Q = CV, where C is a constant called capacitance of the conductor. C = Q ⁄ V For parallel plate capacitor C = i.e. the capacitance depends only on geometrical factors namely, the plate area and plate separation. Spherical Capacitor Isolated Capacitor An isolated sphere can be thought of as a capacitor where other plate is at infinity. R1 = R and R2 = ∞
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 IIT JEE Study Material -  Electrostatics - Capacitor & Combinations

CAPACITORS   CapacitanceIf Q is the charge given to a conductor and V is the potential to which it is raised by this amount of charge, then it is found that Q ∝V. Or Q = CV, where C is a constant called capacitance of the conductor.    C =  Q ⁄ V

For parallel plate capacitor C = 

i.e. the capacitance depends only on geometrical factors namely, the plate area and plate separation.

Spherical Capacitor

Isolated Capacitor

An isolated sphere can be thought of as a capacitor where other plate is at infinity.

    R1 = R     and R2 = ∞

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• Combination of Capacitors

Series Combination: In series combination, each capacitors has equal charge for any value of capacitance. Equivalent capacitance C is given by

 

Parallel Combination:

In parallel combination the potential differences of the capacitor connected in parallel are equal for any of capacitor.  Equivalent capacitance is given by

    C = C1 + C2 + C3

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Example 10.    In the above circuit, find the potential difference across AB.

 

Solution:    Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1.

 

DielectricSubstances having polar atoms/molecules intrinsically or being polarized are called dielectrics.

Polar Dielectric

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Substances which have polar atoms/molecules intrinsically but are randomly arranged. On application of external electric field they get polarized parallel to the external electric field.

Net electric field inside dielectric.

     

      

Capacitance of a parallel plate capacitor having dielectric slabs in series.

 

 Energy stored in a Capacitor

 

Energy supplied by Source

Energy lost in form of heat =

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Energy density of the electric field=

 

Force on a Dielectric in a Capacitor

 

Example 11. A 4μf capacitor is charged to 150 V and another 6μf capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.

Solution :    4μf charged to 150 V would have q1 = C1V1 = 600μC    6μf charged to 200 V would have q2 = C2V2 = 1200μC    After connecting them across each other, they will have a common potential difference V.

    Charges will readjust as q1’ and q2’

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 Electric Field Intensity

It is the force experienced by a unit positive charge placed at a point in an electric field.

Due to a point charge

 

Due to linear distribution of charge

        λ = linear charge density of rod

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(i)    At a point on its axis.

 

(ii)    At a point on the line perpendicular to one end       

Due to ring of uniform charge distribution

At a point on its axis

 

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Due to uniformly charged disc.

At a point on its axis.

Thin spherical shell

 

1.    Non conducting solid sphere having uniform volume distribution of charge(i)    Outside point

(ii)    Inside point    

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           q = total charge

           ρ = volume density of charge

 

Cylindrical Conductor of Infinite length

Outside point

Inside point                    λ = linear density of charge

    E = 0        (as charges reside only on the surface)Non-conducting cylinder having uniform volume density of charge

 

Infinite plane sheet of charge

 

 λ = surface charge density

Two oppositely charged sheets (Infinite)

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(i)    Electric field at points outside the charged sheets    EP = ER = 0

(ii)    Electric field at point in between the charged sheets

Example 3.    Find electric field intensity due to long uniformly charged wire.     (charge per unit length is λ)

 

 

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Example 4.    What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`λ'? The point is separated from the nearer end by a.

 

Solution :    Consider an element, dx at a distance x from the point, P, where we seek to find the electric field. The elemental charge, dq =λdx

 

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Example 5.    A ring shaped conductor with radius R carries a total charge q uniformly distributed around it as shown in the figure. Find the electric field at a point P that lies on the axis of the ring at a distance x from its centre.

 

Solution:    Consider a differential element of the ring of length ds. Charge on this element is

 

This element sets up a differential electric field d  at point P.The resultant field  E¯ at P is found by integrating the effects of all the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d E¯ parallel to this axis contributes to the final result.

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To find the total x-component Ex of the field at P,  we integrate this expression over all segment of the ring.

 

 The integral is simply the circumference of the ring = 2πR

 

As q is positive charge, field is directed away from the centre of the ring, along its axis.

 DIPOLETwo equal and opposite charges separated by a small distance constitute an electric dipole

 

 

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 COULOMB’S LAW

The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, In free space

 

In material medium

 

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Where ε = ε0     εr        εr =  relative permittivity of the mediumin vector form

1.    This is a fundamental law and is based on physical observation2.    The force is an action - reaction pair.3.    The direction of force is always along the line joining the two charges.4.    Electrical force between two point charges is independent of presence or absence of other charges.

Example 1.     A particle ‘A’ having a charge of 2 × 10-6C and a mass of 100g is fixed at the bottom of a smooth inclined plane of inclination 30°. Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium ?

Solution :    First of all draw the F.B.D. of the masses.            For equilibrium ∑F = 0

            N = mg cos30°

 

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Electric Field Intensity    It is the force experienced by a unit positive charge placed at a point in an electric field

                    N/Coul or Volt/m

Example 2.:    Two particles A and B having charges 8 x10-6 C and –2 x10-6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?

Solution :     As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle

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should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B

    Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q

 

Superposition principle    Force experienced by a given charge in the field of a number of point charges is the vector sum of all the forces.

Lines of Force

•    Lines of force originate from positive charges and terminate on negative charges

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•    Lines of force originate or terminate perpendicular to the surface•    Tangent to the lines of force at any point gives the direction of the electric field.

•    Lines of force do not intersect.

 Electric Potential

Amount of work done in moving unit positive charge from infinity up to the point under consideration against the field of a given charge qIn other words, it is negative of the work done by the internal forces

 

It is a scalar quantity

(i)    Potential at a point due to several charges

(ii)    Potential of a point due to an electric dipole

 

Potential due to a uniformly charged disc

(i)    At a point on its axis

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σ = surface charge density on disc

(ii)    At the centre of the disc

Potential at the edge of a uniformly charged disc

 

Potential at the apex of a cone having charge Q distributed uniformly on its curved surface and having slanting length L

 

Potential of a conducting sphere of radius R at a distance r from the centre

 

Potential of a dielectric sphere

 

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Torque on a dipole in presence of external electric fieldTorque = qE. 2a sin θ

τ¯= PE sin θ

Potential Energy U =  -P¯.E¯

Example  8.     Two conducting spheres having radii a and b charged to q1 & q2respectively.  Find the potential difference between 1 & 2 ?

 

Solution :         The potential on the surface of the sphere 1 is given by

 

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Electric Potential Energy

The electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity in order to form the system. For point charges q1 and q2  separated by a distance r12, Electric potential energy of the system q1 and q2 is given by

 

For three particle system q1, q2 and q3

 

We can define electric potential (VP) at any point P in a electric field as

 

where Up, is the change in electric potential energy in  bringing the test charge q0from infinity to point P

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Example 9.    Determine the interaction energy of the point charges of the following set- up

 

 

GAUSS THEOREM

The flux of an electric field E¯  through an arbitrary closed surface S is equal to the algebraic sum of the charge enclosed by this surface divided by ε0.

 

Example 1.    Figure shows a section of an infinite rod of charge having linear charge density λ which is constant for all points on the line. Find electric field E at a distance r from the line.     

Solution :    From symmetry, E¯ due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis.

 

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The direction of E¯ is radially outward for a line of positive charge.

Example 2.    Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively.

Solution:    The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.     

Now, apply Gauss’s law to a spherical Gaussian surface of radius r ( r > R for point A)

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 IIT JEE Study Material -  Wave Optics - Wave Nature of LightWAVE NATURE OF LIGHT

Huygens’ Wave Theory

(i)     Each point on a wave front acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium. 

(ii)     The envelope or the locus of these wavelets in the forward direction gives the position of new wavefront at any subsequent time.

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A surface on which the wave disturbance is in the same phase at all points is called a wave front. 

Wave optics involves effects that depend on the wave nature of light. In fact, it is the results of interference and diffraction that prove that light behaves as a wave rather than a stream of particles (as Newton believed).  

Like other waves, light waves are also associated with a disturbance, which one consists of oscillating electric and magnetic field. The electric field associated with a plane wave propagating along the x-direction can be expressed in the form: 

E→ = E→ o[sin ( λt - kx + θo) ] 

Where λ, k and θo bearing their usual meanings.

Points to remember regarding Interference 

When two waves with amplitude A1 and A2 superimpose at a point, the amplitude of resultant wave is given by

        A = A12 + A2

2 + 2A1A2 cosΦ

Where θ is the phase difference between the two waves at that point.

Intensity (I) = 1/2µoC  Eo2. C = speed of light, E0 = electric field

amplitude

Intensity (I) = I1 + I2 + 2√I1I2 cosθ.

Hence for I to be constant, must be constant.

When θ changes randomly with time, the intensity = I1 + I2.

When θ does not change with time, we get an intensity pattern and the sources are said to be coherent. Coherent sources

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have a constant phase relationship i.e. one that does not change with time.

The intensity at a point becomes a maximum when θ = 2n (n = 0, 1, 2 …) and there is constructive interference.

If θ = (2n - 1) there is destructive interference. (Here n is a non-negative integer)

 

Determination of Phase Difference 

The phase difference between two waves at a point will depend upon 

(a)    The difference in path lengths of the two waves from their respective sources.

(b)    The refractive index of the medium

(c)    Initial phase difference, between the sources, if any.

(d)    Reflections, if any, in the path followed by waves.

 

In case of light waves, the phase difference on account of path difference

= [ Optical path difference / λ ] x 2π  = [ µ(Geometrical path difference) / λ ] 2π  

Where λ is the wavelength in free space. 

In case of reflection, the reflected disturbance differs in phase by λ with respect to the incident one if the wave is incident on a denser medium from a rarer medium. No such change of phase

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occurs when the wave is reflected in going from a denser medium to a rarer medium.

 YOUNG’S DOUBLE SLIT EXPERIMENT

 

A train of plane light waves is incident on a barrier containing two narrow slits separated by a distance’d’. The widths of the slits are small compared with wavelength of the light used, so that interference occurs in the region where the light from S1 overlaps that from S2.

A series of alternately bright and dark bands can be observed on a screen placed in this region of overlap.

The variation in light intensity along the screen near the centre O shown in the figure

Now consider a point P on the screen. The phase difference between the waves at P is θ, where

θ= 2π/λ ΔPo

(where ΔPo is optical path difference, ΔPo=ΔPg; ΔPg  being the geometrical

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path difference.)

 

= 2π/λ  [ S2P - S1P ] (here λ = 1 in air)

As     As, D >> d,

S2P - S1P ≈  λ d sinθ

sin θλ ≈ tanθ( = y/D).

[for very small θ]

 Thus, θ = 2π/λ (dy/D)

For constructive interference,

θ = 2nλ    (n = 0, 1, 2...)

⇒  2π/λ (dy/D) = 2nπ     ⇒ y = n λD/d

Similarly for destructive interference,

y = (2n - 1) λD/2d  (n = 1, 2 ...)

 

Fringe Width W 

It is the separation of two consecutive maxima or two consecutive minima.

Near the centre O [where θ is very small],

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W = yn+1 – yn [yn gives the position of nth maxima on screen]

 = λD/d

 

Intensity Variation on Screen

If A and Io represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position θ as in above figure, is given by

 

I = Io + Io + 2√Io2 cosθ,

When θ = 2π(dsinθ)/ λ = 4Io cos2 Φ/2

 

Illustration 1:

A beam of light consisting of two wavelengths 6500 oA and 5200 oA is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.

(a)    Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 oA.

(b)    What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide?

 

Solution:

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(i) y3 = n. Dλ/d = 3 x 1.2m x 6500 x 10-10m / 2 x 10-3m  = 0.12cm

Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å.

(ii) m x 6500Ao x D/d = n x 5200Ao x D/d ⇒ m/n = 5200/6500 = 4/5

Least distance = y4 = 4.D (6500Ao)/d = 4 x 6500 x 10-10 x 1.2/ 2 x 10-3m = 0.16cm

 

Displacement of Fringes

When a film of thickness’t’ and refractive index 'm' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes.

 

Optical path difference at 

P = S2P - [S1P+ µt - t] = S2P - S1P - (µ - 1)t = y.d/D - (µ - 1)

⇒  nth fringe is shifted by  Δy = D(µ-1)t/d = w/λ (µ-1)t

       

Illustration:

Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10 -5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet?

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 Solution:

As derived earlier, the total fringe shift = w/λ (µ-1)t .

As each fringe width = w,

The number of fringes that will shift = total fringe shift/fring width

(w/λ(µ-1)t)/w = (µ-1)t/λ = (1.6-1) x 1.8 x 10-5m / 600 x 10-9 = 18

  Illustration:

In the YDSE conducted with white light (4000Å-7000Å), consider two points P1 and P2 on the screen at y1=0.2mm and y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points.

   Solution:

The optical path difference at P1 is

p1 =dy1/d = ( 10/4000) (0.2) = 5 x 10-4mm = 5000Å

In the visible range 4000 - 7000Å

n1 = 5000/4000 = 1.25  and n2 = 5000/7000 = 0.714

The only integer between 0.714 and 1.25 is 1

∴ The wavelength which forms maxima at P is    l = 5000Å

For the point P2,

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p2 = dy2/D = ( 10/4000) (1.6) = 4 x 10-3mm = 40000Å

Here n1 = 40000/4000 = 10 and n2 = 40000/7000 = 5.71

The integers between 5.71 and 10 are 6, 7, 8, 9 and 10

∴The wavelengths which form maxima at P2 are

λ1 = 4000 Å                for   n = 10

λ2 = 4444 Å                for   n = 9

λ3 = 5000 Å                for   n = 8

λ4 = 5714 Å                for   n = 7

λ5 = 6666 Å                for   n = 6

 

INTERFERENCE BY THIN FILMA ray of light incident on a thin film of thickness ‘t’ gets partially reflected and refracted at A at surface I and thereafter it gets reflected and refracted at B of surface II. The rays after emerging in the first medium interfere. Now the inference will depend upon the path difference between AD and ABC, as beyond CD path difference is zero

                                                  

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Putting (IV) and (V) in (III) we obtain

    Δ x = 2 μt cosr

AD as reflected at a denser medium it suffers an additional path difference λ/2Total path difference the taken place is

2 μ t cosr –  λ/2

For constructive interference

2μt cos r - λ/2 = μn

>> 2 μt cosr = nλ +λ /2 = (2n + 1) λ/2 maxima

For normal Incidence r =0 >> 2 μt = (2n+1) λ/2 = μn

2 μt cosr - λ/2 = (2n+1) λ/2 >> 2 μt cosr = nλ

For normal incidence 2 μt = nλ

Illustration 6:    White light is incident normally on a glass plate of thickness 0.50 x 10-6 m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate?

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Solution :    The light of wavelength λ is strongly reflected if the light rays reflected are interfering constructively.    As we know the condition for constructive interference

DIFFRACTION

a)    Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path.b)    In Fresnel class of diffraction, the source and/or screen are at a finite

distance from the aperture.   

  

c)    In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhoffer is a special case of Fresnel diffraction

Single Slit Fraunhoffer DiffractionIn order to find  the intensity at point P on the screen as shown in the figure the slit of width 'a' is divided into N parallel strips of width Δx. Each strip then acts as a radiator of Huygen's wavelets and produces a characteristic wave disturbance at P, whose position on the screen for a particular arrangement of apparatus can be described by the angle θ

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The amplitudes ΔEo of the wave disturbances at P from the various strips may be taken as equal if θis not too large.The intensity is proportional to the square of the amplitude. If Im represents the intensity at O, its value at P

is 

 

The concept of diffraction is also useful in deciding the resolving power of optical instruments.

Illustration 7:     Light of wavelength 6 λ 10-5cm falls on a screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum?

Solution:       Here n = 1,     = 6  10-5 cm. 

            Distance of screen from slit = 100 cm.

                        Distance of first minimum from central maxima = 0.1 cm.

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 We know that   a sin= n

                                    a = λ/θ1 = 0.06 cm.

 

 

Illustration 9:            A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment(i)  Find the distance of the third fringe on the screen from the central maximum for the

wavelength 6500A°.(ii)       What is the least distance from the central maximum where the bright fringes due

to both wavelength coincide?(iii)      The distance between the slits is 2mm and the distance between the plane of the

slits and screen is 120cm. What is the fringe width for  = 6500A°?

 

Solution:   

 POLARISATION

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Polarisation of two interfering wave must be same state of polarisation or two source of light should be unpolarised.

BREWSTER LAW

 According to this law when unpolarised light is incident at polarising angle (i) on an interface separating a rarer medium from a denser medium, of refractive index μ as shown in Fig., below such that    μ = tan i

 then light reflected in the rarer medium is completely polarised. Reflected and refractive rays are perpendicular to each other.

REDUCTION IN INTENSITY

Intensity of polarised light is 50% of that of the unpolarised light, i.e.,

     Ip = Iu / 2

where  Ip   = Intensity of polarised light and             Iu   = Intensity of unpolarised light.

 Questions and Solutions part 1

1.    In Young's double slit experiment the light emited from source has l = 6.5 × 10–7 m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be     (a) 3.2 mm    (b) 1.63 mm    (c) 0.585 mm        (d) 2.31 mm

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2.    Light is incident normally on a diffraction grating through which first order diffraction is seen at 32o. The second order diffraction will be seen at     (a) 84o        (b) 48o

    (c) 64o        (d) None of these

Which is not possible. Hence there is no second order diffraction.

3.    In Young's double slit experiment, if the widths of the slit are in the ratio 4:9, ratio of intensity of maxima to intensity of minima will be     (a) 25:1    (b) 9:4    (c) 3:2        (d) 81:16

 Solution-    (a)    As ratio of slit widths = Ratio of intensities

 

4.    A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 550 Å is     (a) 616 Å    (b) 1833 Å     (c) 5500 Å        (d) 6160 Å

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5.    Angular width of a central max. is 30o when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx.     (a) 12 × 10–6 m    (b) 12 × 10–7 m    (c) 12 × 10–8 m    (d) 12 × 10–9 m

 

6.    Light of wavelength 6000 12 × 10–6 m is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then distance between slit and screen will be     (a) 1.0 m    (b) 1.5 m    (c) 2.0 m        (d) 2.3 m                                 

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8.    Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If radius of the sun is 7 × 108 m then time period of rotation of the sun will be       (a) 30 days    (b) 365 days    (c) 24 hours        (d) 25 days.                                 

                               

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10.    If light with wavelength 0.50 mm falls on a slit of width 10 mm and at an angle =  30o to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at      (a) 33.4o         (b) 26.8o

    (c) 39.8o        (d) None of these

 Questions and Solutions part 2

11.    Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. Wavelength of light used is      (a) 3500 Å    (b) 4200 Å    (c) 4700 Å        (d) 6000 Å

 

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12.    A parallel beam of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is 52 then thickness of the film for the reflected light to be coloured yellow (λ = 6000 Å) most intensively must be       (a) 14 (2n + 1) μm        (b) 1.4 (2n + 1) μm    (c) 0.14 (2n + 1) μm        (d) 142 (2n + 1) μm

13.    A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be     (a) 1.1    (b) 1.2    (c) 1.3        (d) 1.5

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14.    In Young's double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thichness 1.964 mm is introduced in the path of one of the two waves. If now mica sheet is removed and distance between slit and screen is doubled, distance between successive max. or min. remains unchanged. The wavelength of the monochromatic light used in the experiment is     (a) 4000 Å    (b) 5500 Å    (c) 5892 Å        (d) 6071 Å

15.    In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If screen is moved by 5 × 10–2 m towards the slits, then change in fringe width is 3 × 10–5 m. If the distance between slits is 10–3 m then wavelength of the light used will be     (a) 4000 Å    (b) 6000 Å    (c) 5890 Å        (d) 8000 Å

16.    White light is used to illuminate two slits in Young's double slit experiment. Separation between the slits is b and the screen is at a distance d(>>b) from the slits. Then wavelengths missing at a point on the screen directly in front of one of the slit are 

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17.    In a Young's double slit experiment, angula width of a fringe formed on a distant screen is 0.1o. If wavelength of light used is 6000 Å , then distance between the slits will be     (a) 0.241 mm    (b) 0.344 mm    (c) 0.519 mm        (d) 0.413 mm

 

18.    Monochromatic light of wavelength 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. Interference pattern is seen on the screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10–6 m and refractive index 1.5 is laced between one of the slits and the screen. If intensity in the absence of plate was Io then new intensity at the centre of the screen will be 

 

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19.       In a Young's interference experimental arrangement incident yellow light is composed of two wavelength 5890 Å and 5895 Å. between the slits is 1 mm and the screen is placed 1 m away. Order upto which fringes can be seen on the screen will be            (a) 384               (b) 486            (c) 512                        (d) 589

 

20.    Ratio of intensities between a point A and that of central fringe is 0.853. Then path difference between two waves at point A will be 

Questions and Solutions part 3

1.    When light is incident on a soap film of thickness 5 × 10–5 cm, wavelength reflected maximum in the visible region is 5320 Å. Refractive index of the film will be    (a) 1.22    (b) 1.33    (c) 1.51        (d) 1.83.

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2.    A ray of unpolarised light is incident on a glass plate of refractive index 1.54 at polarising angle, then angle of refraction is    (a) 33    (b) 44    (c) 57        (d) 90

 

3.    Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be         (a) 0         (b) 2I        (c) 3I        (d) 4I.

Solution - (a) 

    I1 = I2 = I = a2 

    Imax. = (a + a)2 + (2a)2 = 4a2 = 4I 

    Imin. = (a - a)2 = 0

4.    First diffraction minima due to a single slit of width 1.0 x 10–5 cm is at 300. Then wavelength of light used is           a. 400 Å   b.   500Å    c.   600Å    d.    700Å

 

5.    Light of wavelength   is normally incident on a slit. Angular position of second minimum from central maximum us 300. Width of the slit should be   

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    (a) 12 x 10–5 cm        (b) 18 x 10–5 cm        (c) 24 x 10–5 cm        (d) 36 x 10–5 cm

6.    A slit 5 cm wide when irradiated by waves of wavelength 10 mm results in the angular spread of the central maxima on either side of incident light by about  

      (a) 1/2 radian         (b) 1/4 radian        (c) 3 radian        (d) 1/5 radian.                                  Solution - (d)     Angular spread on either side is given by           θ = λ/a = λ/5 radians

7.    In Young's double slit experiment, ten slits separated by a distance of 1 mm are illuminated by a monochromatic light source of wavelength 5 x 10–7 m. If the distance between slit and screen is 2 metre, then separation of bright lines in the interference pattern will be           (a) 0.5 mm        (b) 1.0 mm    (c) 1.5 mm        (d) 1.75 mm

 

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