UNIT-1. Physical Optics and Lasers 1.1 Interference: 1. A soap film (n=1.33) in air is 320 nm thick. If it is illuminated with white light at normal incidence, what color will appear to be in reflected light? [June 2011, 3 marks] Solution: The condition for maxima is 2 µ t cos r = (2n-1) λ /2 where n =1, 2, 3, … Here, µ=1.33, t=320 nm, r=00, Now, λ= By substituting the values of n=1, 2, 3 … we get series of wavelengths which shall be predominately reflected by the film For n=1, λ1=1702 nm; For n=2, λ2= 567 nm; For n=3, λ3= 340 nm Out of these wavelengths only 567nm lies within the visible region. 2. In Newton’s rings experiment, the diameter of the 10th ring changes from 1.40 to 1.27cm, when a liquid is introduced between the lens and the glass plate. Calculate the refractive index of the liquid. [Dec-2008, 3marks] Solution: Given that, Dn(Air)=1.4 cm and Dn(Liquid)=1.27 cm The refractive index of the liquid is given by µ= = =1.215 3. Newton’s rings are formed with reflected light of wavelength = 5 895 1 with a liquid th between the plane and the curved surface. The diameter of the 5 dark ring is 0.3cm and the radius of curvature of the curved surface is 100cm. Calculate the refractive index of the liquid [May/June-2008, 3marks] Solution: Given that, Wavelength of reflected light, λ=5.895X10-5cm Diameter of 5th dark ring, D5=0.3cm Radius of curvature of the curved surface, R=100cm Refractive index of the liquid, µ=? The expression for diameter of 5th Newton’s dark ring with a liquid between the plane and the curved surface is given by, = µ=

=

= 1.31

4. In Newton’s rings experiment the diameter of the 14th and 12th dark rings are 4mm and 7mm respectively. Find the diameter of the 20th dark ring. Solution: Given D12=Dn+p=0.7 cm and D4=Dn=0.4 cm n+p=12; n=4 Therefore p=8 We have Dn+p2 - Dn2 =4 λ 0.72 – 0.42 =4x8xλ => λ = 1 3 Diameter of nth dark ring is given by Dn= 2√ λ D20 = √2 1 3= 9 6 Therefore D20=0.906cm 5. Newton’s rings formed by sodium light between glass plate and a convex lens are viewed normally. Find the order of the dark ring which will have double the diameter of that of 30th ring. [June 2012, 2 marks] Solution: Given n=30;

λ =5890A0;

n+p=?

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Diameter of nth dark ring is given by Dn=√4 λ =√4 12 Given Dn+p=2Dn=2x√12 λ We have

=4

3

λ =√12 λ

=

12

Dn+p2 - Dn2 =4 λ

(4-1)120 λ =4 λ

P=90 and hence n+p=120 is the required order of dark ring.

6. Newton’s rings are observed in reflected light of wavelength 5900A0. The diameter of 10th dark ring is 0.50cm. Find the radius of curvature of the lens and the thickness of the air film.[Dec2007, 3marks] Solution: Given n=10th ring; D10=0.5cm; r10=0.25cm; λ=5900A0 We know that radius of nth Newton’s dark ring is given by rn = √ λ Therefore, rn2 = λ => R= rn2/ λ Radius of curvature of the lens R= = 105.9cm Thickness of the Air film t= rn2/2 = =2.945X10-4cm 7. If the diameter of two consecutive Newton’s rings in reflected light of wavelength 5890A0 are 2.0 and 2.02 cm respectively. Calculate radius of curvature of the lens [June-2013, 2marks] Solution: Given Dn=2cm; Dn+1=2.02cm λ=5890A0 We know that

Dn+p2 - Dn2 =4 λ

Dn+12 - Dn2 =4λ 2.022 - 22 =4589

1

R=341 cm 1.2 Diffraction: 1. How many orders will be observed by a grating having 4000 lines per cm if it is illuminated by visible light in the range 4000 A0 and 7000 A0? [Jan 2012, 3 marks] Solution: Given N=4000 lines/cm, For

=4

=4

1

; formula

=

=

=

=

For

=7

=4

1

; formula

=

=

=

=

The order of the spectrum varies from 3 to 6 depending upon the wavelength of the visible light.

2. Calculate the minimum number of lines per cm in a 2.5 cm wide grating which will just resolve the sodium lines (5890 A0 and 5896 A0) in the first order spectrum. [June 2012, 2 marks] Solution: Given = 589 = 589 1 ; = 5896 = 5896 1 = = 5896 1 589 1 =6 1 = 589 1 = 981 66 1 6 1 herefore for proper resolution, the minimum number of lines in grating will be 982 =

=

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3. How many orders will be visible if the wavelength of the incident radiation deviated at an angle of 300 is 5893 A0 and the number of lines on the grating is 2540 per inch? [Jan 2013, 3 marks] Solution: Given = 3 = 5893 = 5893 1 N=2540 lines/inch=1000 lines/cm We know that

=

=

=

= 8 48

Therefore 8 orders will be visible. 1.3 Polarization: 1. A plane polarized light of wavelength 6000 A0 is incident on a thin quartz plate cut with faces parallel to the optic axis. Calculate the minimum thickness of the plate which introduces a phase difference 600 between the ordinary and extraordinary rays. [June 2010, 3 marks] Solution: Assume = 1 544 = 1 553 and = 6 Since, a phase difference of 3600 corresponds to a path difference of , a phase difference of 600 corresponds to a path difference of /6. 6 1 = = = = 6 6 6 1 553 1 544 2. Two Nicols have parallel polarizing directions so that the intensity of transmitted light is maximum. Trough what angle must either Nicol be turned if intensity is to drop by one-fourth of its maximum value? [June 2011, 2 marks] Solution:- According to Malus Law, the intensity of transmitted light through the analyzer is given by = ; where Io is the intensity of incident polarized light on the analyser. When two Nicols have parallel polarizing direction the intensity of transmitted light is maximum, i.e., = = To get one fourth of the maximum value, either Nicol be turned through an angle θ. For this the Malus law can be written as Therefore

=

=

=

=

(

)=

6

, 12

3. Determine the specific rotation of the sugar solution, if the plane of polarization is turned through 15o. The length of the tube containing 20% of sugar is 20cm. (Dec-2013, 2) Solution: Given = 15 , = 2 =2 , =2 = 2 / We know that specific rotation

= =

=

15 2

2

= 37 5

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UNIT-2. Modern Physics 2.2 Wave mechanics: 1. The energy of an electron contained to move in a one-dimensional box of length 4.0 A0 is 9.664 X 10-17 J. Find out the order of excited state. [Jan 2012, 2 marks] Solution: The possible energies of a particle in a 1-D box of size L is given by =

8

=

9 664

1

8 91 1 6 625 1 = 16

4

1

= = 256

2.3 Electromagnetic theory: 1. Calculate the value of pointing vector at the surface of the sun if the power radiated by the sun is 3.8 X 1026 W and its radius is 7 X 108 m. [June 2011, 2 marks] Solution: Given P=3.8 X 1026 W and r=7 X 108 m 38 1 = = = = 6 174 1 / 4 4 3 14 7 1

UNIT-3. Condensed matter Physics 3.1 Crystallography: 1. The first order diffraction is found to occur at a glancing angle of 90. Calculate the wavelength of X-rays and the glancing angle of second order diffraction if the spacing between the adjacent planes is 2.51 A0. [June 2011, 3 marks] Solution: Given = 1, = 9 = 2 51 = 2 51 1 Formula 2 sin = = 2 2 51 1 sin 9 = 7853 =2

= sin

( ) = sin

( ) = sin

(

) = 18 2

3.3 Semiconductors: 1. For an intrinsic semiconductor having band gap Eg= 0.7eV, calculate the density of holes and electrons at room temperature (27 0C). Given K=1.38 X 10-23 J/K and h=6.62X10-34 J. [June 2011, 3 marks] Solution:

= 2(

Assuming

=

)

exp (

)

= = 2(

2

) exp ( 2

)

B.E. ¼ - Engineering Physics - O.U. Question Papers – Numerical Problems with solutions Page 4 of 5

2 = 2(

22

91

1 1 38 7 6 62 1 =

1

3

) exp ( 2

7 16 1 38 1

1 3

)

/

2. Mobilities of electrons and holes in a sample of intrinsic Ge at 300 K are 0.36 m 2V-1S-1 and 0.17 m2V-1S-1 respectively. If the resistivity of the specimen is 2.12 ohm –m, compute the intrinsic concentration of carriers for Ge. Where me*=0.5 m0 and mh*=0.37 m0. [Jan 2012, 3 marks] Solution: Resistivity

=

= 1

=

=

1 2 12

16

1

36

17

=

/

3. The intrinsic carrier density at 300 K in silicon is 1.62 X 1016 /m3. If the electron and hole Mobilities are 0.13 and 0.06 m2V-1S-1 respectively. Calculate the conductivity of intrinsic silicon. (June-2013, 3) Solution: Conductivity = = 1 62 1 16 1 13 6 = 49 1 /

UNIT-4. Material Science 4.3. Superconductivity: 1. The superconducting transition temperature of a metal is 7.26K. The critical field at 0 K is 64 X 103 A/m. Calculate the critical field at 5K. [Jan 2013, 3 marks] Solution: Critical Field

=

*

+=

*

+=

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