Assignment Questions and Solutions

THE NATURE AND PROPAGATION OF LIGHT

Question 33.9:

Answer:

Question 33.14:

Answer:

1

Question 33.21:

Answer:

Question 33.30:

Answer:

2

GEOMETRIC OPTICS

Question 34.4:

Answer:

Question 34.28:

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3

Question 34.53:

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4

INTERFERENCE

Question 35.10:

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Question 35.20:

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5

Question 35.37:Light containing two wavelengths 1 and 2 falls normally, giving Newtons rings. If the (n-1)th dark ringdue to 1 coincide with the nth dark ring due to 2, prove that the radius of the nth dark ring due to 2

is 1 2R(1 2) .Answer: The diameter of (n-1)th dark ring due to 1 is given byDn1=4 (n1) 1 R

The diameter of nth the dark ring due to 2 is given byDn=4 n 2R

As per the question, Dn-1 = Dn

Hence, 4 (n1) 1R=4 n 2R

(n1 ) 1=n 2

Or

n(12)= 1 ; n=1

(1 2)

Now

Radius of the nth dark ring is rn=n 2R= 1 2R( 1 2)

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DIFFRACTION

Question 36.2:

A He-Ne laser of 6328 illuminates a 0.4 mm wide slit. Find the width of the central maxima on ascreen kept at a distance of 1.6m.Answer:

= 6328 = 6328 x 10-10 m; a=0.4mm=0.0004m; f=1.6 m

/a=y/f

Or y= (f/a)

i.e. y=(6328 x 10-10x1.6)/0.0004=2.53x10-3 m

Hence Full width of the central maxima is = 2y=2x2.53x10-3=5.06 x10-3 m =5.06mm

Question 36.15:In Fraunhofer, diffraction due to narrow slit of width 0.1 mm, the first maxima lies 15.8 mm on eitherside of the central maxima on a screen kept at a distance of 2.5 m from the slit. Calculate thewavelength of the light used.

Answer:

a=0.1mm=0.0001m; f=2.5 m, y1= 15.8mm=15.8x10-3 m

Here first maxima means the first secondary maxima. The condition of first secondary maxima in

single slit is a sin = (m+1/2) ; Here m = 1; hence (m+1/2) = 3/2=1.5

(m+1/2)/a=y1/f; where m = 1

Or = y1a/(m+1/2)f

= (15.8x10-3x0.0001)/(1.5x2.5)= 4.213x10-7 m

= 4213

7

Question 36.30: Parallel beam of the two wavelengths 5890 and 5896 of a sodium vapour lamp fall on a diffraction grating having 6000 lines/cm. Estimate the dispersion produced by the grating of two wavelengths, in the first order spectra

Answer:

1 = 5890 = 5890 x 10-8 cm and 2 = 5896 = 5896 x 10-8 cm

d = (a+b) = 1/6000 = 1.166x10-4 cm

(a+b)sin 1=n 1 where n = 1

1 = sin-1{ 1/(a+b)}= sin-1{5890 x 10-8/(1.166x10-4)}

Hence, 1 =30.33

(a+b)sin 2=n 2 where n = 1

2 = sin-1{ 2/(a+b)}= sin-1{5896 x 10-8/(1.166x10-4)}

Hence, 2 =30.37

Hence the dispersive angle (angular separation) for both the wavelength is

d = 2-1 = 30.37 -30.33

d = 0.04 =6.98x10-4 radians

Dispersive power = d /d = 6.98 x 10-4 radians /(6 x 10-10 m) = 1.16 x 106 radians/m

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degree

degree

degree

degree

RELATIVITY

Question 37.5:Two thousand -mesons are produced at an altitude of 40 km. how many -mesons should an averagereach sea level before they decay if the velocity of mesons is 0.999c and if their mean life is 2.2x10 -6 s.when are they rest? (Assume the -mesons travel vertically down without loss of energy).

Answer:

Given u = 0.999c; N0 = 2000

t0 = 2.2 x 10-6 s

t= t0

(1u2c2 )Hence, t = 4.9205 x 10-5 s =

The time period travels by -mesons from the altitude 40 km to the sea level,

t =s/u=40km/0.999c

t = 13.34 x 10-5 s

Now apply, N=N oet

N=2000 e13 .34 x 105

4 . 92 x105

N = 132

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Question 37.10: Find the proper of a rod, if in the laboratory frame its velocity is u=c/3 and its length l =1.5 m and the angle between the rod and direction of motion is 30.Answer:

u = c/3; l = 1.5 m and = 30

The length of the rod in horizontal direction

l=l0 cos(1u2c2 )Or

l0=l

cos(1u2c2 )Hence,

l0=1.837m

Question 37.18: The density of a stationary body is equal to 0. Find the velocity of an observer for whom the density of the body measures 40 % more.

Answer:

Given, =0+0.40 =1.4 0

Given /0 = 1.4

= 0/[1-(u2/c2)]1/2

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Or 0/ =[1-(u2/c2)]1/2

(1/1.4)2 = 1-(u2/c2)

0.510=1-(u2/c2)

Hence, u 2=( 1-0.5102)c

So, u = 0.699c

Question 37.27:

Answer:

Question 37.33:

Answer:

11

PHOTONS: LIGHT WAVES BEHAVING AS PARTICLESQuestion 38.9:

Answer:

Question 38.17:

Answer:

12

Question 38.23:Electrons in an X-ray tube are accelerated by a potential difference of 15.0 KV. If an electronproduces a photon on impact with the target, what is the minimum wavelength of theresulting X-rays?Answer:

Given, VAC = 15 KV = 15 x 103 V

eVAC = hfmax = hc/

Or = hc/eVAC = (6.626x 10-34 x 3 x 108)/ (1.601 x 10-19 x 15 x 103) = 0.0827 nm

Question 38.28:

Answer:

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PARTICLES BEHAVING AS WAVES

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Question 39.26:Find the kinetic, potential and total energies of the electron in hydrogen atom in the second excitedlevel, and find the wavelength of the photon emitted in the transition from the second excited level tothe ground level.Answer:

So, Kn = 13.6/n2 eV

Here, n = 3 (second excited level)

Hence, Kn= 1.511 eV

So, Un = -27.2/n2 eV

Here, n = 3 (second excited level)

Hence, Un= -3.022 eV

So, Kn = 13.6/n2 eV

Here, n = 3 (second excited level)

Hence, En= -13.6/9 = -1.511 eV

nL = 1 and nU = 3

R = 1.07 x 10-7 m-1

= 105.1 nm

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Question 39.37:

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16

Question 39.42:A body emits radiation of wavelength m=0.5x10-6 m corresponding to its maximum energy. What is itsabsolute temperature?Answer:

Given, m = 0.5x10-6 m

Apply Wien displacement law, is given by

mT = constant (b) = 2.90x 10-3 m.K

Hence, T = 2.90x 10-3 /0.5 x 10-6 = 5800 K

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Answer:

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QUANTUM MECHANICS

Question 40.2:

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Question 40.13:

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Question 40.31:In the case of a particle in a box of length L, divide the region 0 to L into 3 parts, 0 to L/4, L/4to 3L/4 and 3L/4 to L. Calculate the probability of finding the particle in any of these threeregions when it is in the ground state.Answer:For particle in a Box, ground state (n = 1) wave function is , (x) = (2/L)1/2Sin(

P=x1

x2

| (x)|2dx= 2Lx 1x 2

sin2 xLdx

P=[ xL 12 sin 2xL ]x 1x 2

P1=[ xL 12 sin 2xL ]0L/4

P1 =0.25-0-1/2 =0.0908

P2=[ xL 12 sin 2xL ]L/43L/ 4

P2=0.5

P3=[ xL 12 sin 2xL ]3L/ 4L

P3 =0.25+1/2=0.4092

Hence the total probability is P = P1 + P2 + P3 = 0.091+0.5+0.4092 = 1

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xL)

Question 40.35:

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Question 40.42:

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ADDITIONAL NUMERICALS

INTERFERENCEQuestion:

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DIFFRACTIONQuestion:

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RELATIVITY

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PHOTONS: LIGHT WAVES BEHAVING AS PARTICLES

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PARTICLE BEHAVING AS WAVES

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QUANTUM MECHANICS

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