## Pages

### Polarization by reflection

• This simple method of obtaining plane polarized light by reflection was discovered by malus in 1808
• We found that when a beam of light is reflected from the surface of a transparent medium like glass or water, the reflected light is partially polarized and degree of the polarization varies with angle of incidence
• The percentage of polarized light is greatest in reflected beam when light beam is incident on the transparent medium with an incident angle equal to the angle of polarization
• For ordinary glass with refractive index =1.52 ,angle of polarization is 57.50
• Figure below shows the polarization of light by reflection

• we can use a Polaroid as an analyzer to show that reflected light is plane polarized . we rather say that reflected light is partially plane polarized
• the examination of transmitted light shows the variation in intensity indicating that the light is partially polarized
• The vibrations of this plane polarized reflected light are found to be perpendicular to the plane of incidence and therefore ,the reflected light is said to be plane polarized in the plane of incidence

### Kinematics Question

Question:

A ball of mass 100 g is projected vertically upward from the ground with avelocity of 49m/s. At the same time another identical ball is dropped from aheight of 98 m to fall freely along the same path as followed by the firstball. After some time ,the two balls collide and stick together and finallyfall together .

g=9.8m/s2

i is the unit vector along upward direction

1) Find the time at which balls collide in air

a) 2.1 sec

b) 1.2 sec

c)1 sec

d) 2 sec

2) Find the velocity vector of the combined mass just after collision

a) 4.8i

b) 3i

c)-4.9i

d) 4.9i

3) Find the time of the flight of masses

a) 3sec

b) 4 sec

c) 6.53 sec

d) 7 sec

Solution:

a) We will first find where and when the two balls collide. Let us assumethat the balls collide at time t after they have been set into motion. At thisinstant t when two balls collide they are at the same height h from the groundas shown below in the figure.

The height of the first ball after t seconds = 49t-0.5(9.8t2) =4.9t(20-t)

Height of second ball after t secinds = 98 - downwards distance moved by itin t seconds

=98-0.5t2=4.9(20-t2)

therefore, 4.9t(20-t)=4.9(20-t2)

or, 10t-t2=20-t2 or t=2s

b) The ball thus collides 2s after the start of their motion. Theirvelocities at this instance are

ball 1 : v1= (49-98×2)m/s = 29.4 m/s directed upwards=29.4 i

ball 2 : v2=(0+9.8×2)m/s = 19.6 m/s directed downwards=-19.6 i

If v is the velocity of the combined mass of two balls after theystick togather due to their collision then from law of conservation of momentum

200×v=100×29.4-100×19.6

v=4.9m/s upward direction =4.9 i

c) The joint mass thus moves upwards , after collision with a velocity of4.9 m/s. Its height above the ground at this instant is (consider the positionof either of the balls)

(98-0.5×9.8×22)m=78.4m

We can now find the time t' taken by this joined mass of the balls to reachthe ground. For this joined mass we have

u=4.9m/s , s=78.4m , a=-g = -9.8m/s2

-78.4=4.9t'+0.5(-9.8)t2

t'2-t'-16=0

Solving the equation for t' using formula for quadratic equations and leaving out the negative solution we get t'=4.532 s

The joint mass thus takes 4.53 s to fall to the ground. Since the balls collide 2s after they started their motion ,

the total time of flight is (2+4.53) s = 6.53 s

### Question on Bolr atom model

Question :

The ionization energy of a hydrogen like Bohr atom is 4 rydbergs.

(a) What is the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state.

(b) Whet is the radius of the first orbit of this atom ? Given that Bohr radius of hydrogen atom = 5×10-11m and 1 rydberg = 2.2×10-18J.

Solution :

In terms of rydberg constant R , the energy of electron in the n'th orbit of hydrogen like atom is

${E}_{n}=-\frac{R{Z}^{2}}{{n}^{2}}$

where R=2.2×10-18J. The ionization energy of the atom is

${\mathrm{\Delta E}=E}_{\infty }-{E}_{1}=-R{Z}^{2}\left(\frac{1}{\infty }-\frac{1}{{1}^{2}}\right)=R{Z}^{2}$

Given that ΔE=4R. Therefore , 4R=RZ2 or, Z=2

(a) The energy of the radiation emitted when the electron jumps from the first excited state (n=2) to the ground state (n=1) is

${E=E}_{2}-{E}_{1}=-R{Z}^{2}\left(\frac{1}{{2}^{2}}-\frac{1}{{1}^{2}}\right)=\frac{3R{Z}^{2}}{4}=3R=3×2.2×{10}^{-18}=6.6×{10}^{-18}J$ (since Z=2)

Therefore wavelength of the radiation is given by

$\lambda =\frac{\mathrm{hc}}{E}=\frac{6.63×{10}^{-34}×3×{10}^{8}}{6.6×{10}^{-18}}=301Å$

(b) Radius of the first bohr orbit of the given atom is $=\frac{{r}_{1}}{Z}=2.5×{10}^{-11}m$

### Seeback Effect

• Seeback effect was first discovered by Thomas John seaback
• It stated that when two different conductor are joined to form a circuit and the two junctions are held at the different temperature then an emf is developed which results in the flow of the electric current through the circuit.Arrangement is shown as below in figure
• Maginitude of thermo-electric emf depends upon the nature of the two metals and on the temperature difference between terminals

• Seaback effect is reversible i.e, if the hot and cold junctions are reversed the direction of thermoelectric current is alse reversed
• Seaback investigated thermo-electric properties of a large number of metals and arranged them in a series known as thermo-electric series or seaback series and is given as follows
Bi,Ni,Co,Pt,Cu,Mn,Hg,Pb,Sn,Au,Ag,Zn,Cd,Fe,As,Sb,Te
• When any two of these metals in the series is used to form a thermocouple ,the thermo emf is greater when two metals used are farther apart in the circuit
• Figure 1 shows the thermocouple of Cu and Fe.The current in this couple flows from Cu to Fe through the hot junction
• The thermo emf of this couple is only 1.3 milivolt for a temperature difference of 100 C between the hot and cold junction

### Question on electric charge

Question
Two isolated metallic spheres of radii R and 2R are charged such that both these have the same charge densiti σ. The sphares are located far away from each other and connected by a thin conducting wire. Find the new charge density on the bigger wire.
Solution
Initial charges on sphere of radius R and 2R respectively are
Q1= surface area × charge density = 4πR2σ
and Q2=4π(2R)2σ = 16πR2σ
therefore Q1+Q2=4πR2σ+16πR2σ (1)
As the charges are different, their initial potentials are also different. When the spheres are connected by long thin conducting wire, the charges are redistributed untill their potentials become equal. Let Q'1 and Q'2 be the new charges on spheres of radii R and 2R respectively, their common potential is
$V=\frac{1}{4\pi {\epsilon }_{0}}.\frac{{Q\text{'}}_{1}}{R}=\frac{1}{4\pi {\epsilon }_{0}}.\frac{{Q\text{'}}_{2}}{2R}$
which gives
${Q\text{'}}_{1}=\frac{{Q\text{'}}_{2}}{2}$
Therefore,
${Q\text{'}}_{1}+{Q\text{'}}_{2}=\frac{3{Q\text{'}}_{2}}{2}$
From Law of conservation of charge
Q1+Q2=Q'1+Q'2
Using equation 1 and 2 we have
${Q\text{'}}_{2}=\frac{40\pi {R}^{2}\sigma }{3}$
The new charge density on the bigger sphere of radius R is
$\sigma \text{'}=\frac{{Q\text{'}}_{2}}{4\pi {\left(2R\right)}^{2}}=\frac{5\sigma }{6}$