(a) Use Huygens' principle to show the propagation of a plane wavefront from a denser medium to a rarer medium. Hence find the ratio of the speeds of wavefront in the two media.
(b) (i) Why does an unpolarized light incident on a polaroid get linearly polarized ?
(ii) Derive the expression of Brewster's law when unpolarized light passing from a rarer to a denser medium gets polarized on reflection at the interface.
As seen in the fig. above let XY be a surface separating the two media ‘1’ and ‘2’. Let v1 and v2 be the speeds of waves in these media.
A plane wavefront AB in the first medium is incident obliquely on the boundary surface XY and its end A touches the surface at A at time t = 0 while the other end B reaches the surface at point B after time-interval t.
As the wavefront AB advances, it strikes the points between A and B’ of boundary surface.
According to Huygens principle, secondary spherical wavelets emanate from these points, which travel with speed v1 in the first medium and speed v2 in the second medium.
Secondary wavelet starting from A, traverses a distance AA’ = v2t in second medium in time t. In the same time, point of wavefront traverses a distance in first medium and reaches B’, from where the secondary wavelet starts.
So, BB' = v1 t and AA’ = v2t.
Assuming A as centre, we draw a spherical arc of radius AA’ (= v2t) and draw tangent B’A’ on this arc from B’. As the incident wavefront AB advances, the secondary wavelets start from points between A and B’, one after the other and will touch A’B’ simultaneously.
According to Huygens principle A’B’ is the new position of wavefront AB in the second medium. Hence A’B’ will be the refracted wavefront.
Let the angle made by incident wavefront be i and angle made by the refracted wavefront A’B’ be r.
Polarization of light is referred to as restricting the vibration of light in a perpendicular direction perpendicular to the direction of propagation of wave.
The vibration of particles of light which is parallel to the axis of crystal passes through the Polaroid on passing an unpolarized light. All other vibrations are absorbed and that is why intensity of emerging light is reduced.
The plane of vibration here is ABCD, in which the vibrations of the polarized light is confined and the plane KLMN is called plane of polarization. KLMN is perpendicular to the plane of vibration.
Reflected light is totally polarized, when unpolarized light is incident on the glass-air interface at the Brewster angle iB. This is known as Brewster’s law.
The reflected component OB and refracted component OC are mutually perpendicular to each other, when light is incident at Brewster’s angle.
i = iB and r = (900 – iB)
, is the expression for Brewster’s law.
A biconvex lens with its two faces of equal radius of curvature R is made of a transparent medium of refractive index 1. It is kept in contact with a medium of refractive index 2 as shown in the figure.
a) Find the equivalent focal length of the combination.
b) Obtain the condition when this combination acts as a diverging lens.
c) Draw the ray diagram for the case when the object is kept far away from the lens. Point out the nature of the image formed by the system.
a) Using the Len’s maker’s formula, we have
Let feq is the equivalent focal length of the combination, then
For, the combination of lenses to behave as diverging lens, equivalent focal length < 0.
For the combination of lenses will behave as converging lens. An object placed far from the lens will form image at the focus of the lens.
Image formed is real and diminished in nature.
During a thunderstorm the 'live' wire of the transmission line fell down on the ground from the poles in the street. A group of boys, who passed through, noticed it and some of them wanted to place the wire by the side. As they were approaching the wire and trying to lift the cable, Anuj noticed it and immediately pushed them away, thus preventing them from touching the live wire. During pushing some of them got hurt. Anuj took them to a doctor to get them medical aid.
Based on the above paragraph, answer the following questions:
(a) Write the two values which Anuj displayed during the incident.
(b) Why is it that a bird can sit on a suspended 'live' wire without any harm whereas touching it on the ground can give a fatal shock?(c) The electric power from a power plant is set up to a very high voltage before transmitting it to distant consumers. Explain, why.
a) Anuj displayed concern for others lives, presence of mind and a selfless attitude.
b) When the bird perches on a live wire, body becomes charged for a moment and has a same voltage as the live wire. However, no current flows in its body. Body is a poor conductor of electricity as compared to the copper wire. So the electrons do not travel through the bird’s body.
On the other hand, if the bird touches the ground while being in contact with the high voltage live wire, then the electric circuit gets complete and high current flows through the body of the bird to the ground, giving it a fatal shock.
c) Inorder to reduce the loss of power transmission, the power plant is set up at high voltages. Power loss during the transmission is I2R. Therefore, by reducing the value of current, power loss can be minimized. So, the value of the voltage should be kept high.
(a) Define the term 'intensity of radiation' in terms of photon picture of light.
(b) Two monochromatic beams, one red and the other blue, have the same intensity.
In which case:
(i) the number of photons per unit area per second is larger,
(ii) the maximum kinetic energy of the photoelectrons is more? Justify your answer.
a) The number of photons falling per unit area in unit time is defined as the intensity of radiation.
i) Since, the two beams have the same intensity therefore, the number of photons emitted per unit are per unit time is the same.
ii) Maximum Kinetic energy of photoelectrons is given by,
We know that the frequency of blue beam is more than that of the red beam. Therefore, maximum Kinetic energy of the blue beam will be more.
Draw a necessary arrangement for winding of primary and secondary coils in a step-up transformer. State its underlying principle and derive the relation between the primary and secondary voltages in terms of number of primary and secondary turns. Mention the two basic assumptions used in obtaining the above relation.
State any two causes of energy loss in actual transformers.
Underlying principle of a step-up transformer: A transformer which increases the ac voltage is known as a step up transformer.
Working of step-up transformer is based on the principle of mutual inductance and it converts the alternating low voltage to alternating high voltage. The number of turns in the secondary coil is greater than the number of turns in the primary coil.
i.e., NS > NP
Working: When an A.C source is connected to the ends of the primary coil, the current changes continuously in the primary coil. Hence, the magnetic flux which is linked with the secondary coil changes continuously. So, the emf which is developed across the secondary coil is same as that in the primary coil. The emf is induced in the coil as per Faraday’s law.
Let, EP be the alternating emf applied to primary coil and np be the number of turns in it.
Consider as the electric flux associated with it.
Assumption: We assume that there is no leakage of flux so that, the flux linked with each turn of primary coil is same as flux linked with secondary coil.
Two sources of energy loss in the transformer:
Joule Heating – Energy is lost in resistance of primary and secondary windings in the form of heat.
H = I2Rt
Flux leakage - Energy is lost due to coupling of primary and secondary coils not being perfect, i.e., whole of magnetic flux generated in primary coil is not linked with the secondary coil.
Conservation of law of energy is not violated in step-up transformer. When output voltage increases, the output current automatically decreases. Thus, there is no loss of energy.
Two infinitely long straight parallel wires, '1' and '2', carrying steady currents I1 and I2 in the same direction are separated by a distance d. Obtain the expression for the magnetic field due to the wire '1' acting on wire '2'. Hence find out, with the help of a suitable diagram, the magnitude and direction of this force per unit length on wire '2' due to wire '1'. How does the nature of this force changes if the currents are in opposite direction? Use this expression to define the S.I. unit of current.
Consider a straight conductor XY lying in the plane of paper. Consider a point P at a perpendicular distance from the straight conductor.
Magnetic field induction (B) at a point P due to current I passing through conductor XY is given by,
where, are the angles made by point X and Y respectively.
At the centre of the infinite long wire, 1 = 2 = 90o
So, magnetic field is given by,
Magnetic field produced by current I1 at any point on conductor Rs is given by,
Force acting on length l of the conductor RS will be,
An equal force is exerted into the wire PQ by the field of conductor RS which is given by,
Thus, the force is attractive when the current is acting along the same direction. When, current flows in opposite direction, the forces between the two conductors are repulsive.
One Ampere is that value of constant current which when flowing through each of the two parallel uniform long conductors placed in free space at a distance of 1m from each other will attract or repel with a force of 2 10-7 Newton per metre of their length.
a) State Kirchhoff's rules and explain on what basis they are justified.
(b) Two cells of emfs E1 and E2 and internal resistances r1 and r2 are connected in parallel.
Derive the expression for the
(i) Emf and
(ii) internal resistance of a single equivalent cell which can replace this combination.
(a) "The outward electric flux due to charge +Q is independent of the shape and size of the surface which encloses is." Give two reasons to justify this statement.
(b) Two identical circular loops '1' and '2' of radius R each have linear charge densities -and + C/m respectively. The loops are placed coaxially with their centre distance apart. Find the:magnitude and direction of the net electric field at the centre of loop '1'.
i) Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is always zero.
This law is in accordance with law of conservation of charge.
ii) Loop Rule: In a closed loop, the algebraic sum of emfs is equal to the algebraic sum of the products of the resistances and the current flowing through them.
This law is based on the conservation of energy.
b) Consider the circuit,
Here, E1 and E2 are the emf of two cells,
r1 and r2 are the internal resistance of cell,
I1 and I2 current due to two cells.
Terminal potential difference across the first cell is given by,
For the second cell, terminal potential difference will be equal to that across the forst cell.
Let E be the effective emf and r the resultant internal resistance.
Consider, I as the current flowing through the cell.
Now, comparing the equation with V = E –Ir, we have
a) The outward electric flux due to the charge enclosed inside a surface is the number of electric field lines coming out of the surface. Outward flux is independent of the shape and size of the surface because:
i) Number of electric field lines coming out from a closed surface is dependent on charge which does not change with the shape and size of the conductor.
ii) Number of electric lines is independent of the position of the charge inside the closed surface.
b) Magnitude of electric field at any point on the axis of a uniformly charged loop is given by,
Electric field at the centre of loop 1 due to charge present on loop 1 = 0
Electric field at the centre of loop1 due to charge present on loop 2 is given by,
, is the required electric field.
Obtain the relation between the decay constant and half-life of a radioactive sample.
The half-life of a certain radioactive material against decay is 100 days. After how much time, will the undecayed fraction of the material be 6.25%?
In a radioactive sample, number of atoms at any instant is given by,
Now, when t = T, where T is the half-life of the sample.
This implies, half-life of a radioactive substance is inversely proportional to decay constant.
Number of undecayed nuclei left = 6.25 % = 6.25/ 100 = 1/16
Let, t be the required time after which the undecayed fraction of the material will be 6.25%.