Session Ending Exam, 2021-22
Class-XII
Subject –Physics (042)
Max. Marks 35 Max. Time 2 Hrs
General Instructions:
(i) There are 12 questions in all. All questions are compulsory.
(ii) This question paper has three sections: Section A, Section B and Section C.
(iii) Section A contains three questions of two marks each,
Section B contains eight questions of three marks each,
Section C contains one case study-based question of five marks.
(iv) There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the choices in such questions.
(v) You may use log tables if necessary but use of calculator is not allowed.
SECTION – A
1. i) What type of extrinsic semiconductor is formed when
a) germanium is doped with indium? b) silicon is doped with bismuth?
Ans. a) p type semiconductor b) n type semiconductor
ii) In a semiconductor the concentration of electrons is 8×1013 cm-3 and that of holes is 5 x 1012 cm-3. Is it a p-type or n- type semiconductor?
Ans. As concentration of electrons more than concentration of holes, the given extrinsic semiconductor is n type.
2. With the help of Rydberg formula find the wavelength of first line of Pfund series.
OR
When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5V0. When the same surface is illuminated with the light of wavelength 3λ, the stopping potential is V0. What is the work function of the metallic surface?
3. i) Explain why elemental semiconductor cannot be used to make visible LEDs. ii) In the following diagram, which bulb out of B1 and B2 will glow and Why?
(i) Elemental semiconductor’s band gap such that electromagnetic emission is in Infrared region.
ii) In the diagram, diode D1 is in forward bias and diode D2 is in reverse bias, so current will flow through D1 and D2 will not allow flowing the current, so bulb B1 will glow and bulb B2 will not glow.
SECTION – B
4. When an electron in hydrogen atom jumps from the third excited state to the ground state, how would the de Broglie wavelength associated with the electron change? Justify your answer.
5. A student wants to use two p-n junction diodes to convert alternating current into direct current. Draw the labelled circuit diagram he/she would use and explain how it works.
At any instant the voltage at A (input voltage of diode 1) and end B (input voltage of diode 2) of the secondary with respect to the centre tap will be out of phase. Suppose during a positive half cycle of Ac input, the end A is positive and end B is negative with respect to the centre tap.
Then diode 1 gets forward biased (allows the flow of current) and diode 2 gets reversed biased (does not allow the flow of current).
Hence the current flows through the diode 1 towards the centre tap along the path AXY as shown in the above diagram. Similarly, during the negative half cycle of Ac input the end B becomes positive and end A becomes negative.
Hence diode 1 gets reversed biased and diode 2 gets forward biased resulting in the current to flow from diode 2 to the centre tap along the path BXY as shown in the above diagram. As during both the half cycles the input a.c. the current through the load resistor will flow from X to Y and will keep on pulsating.
6. i) Draw a graph showing variation of potential energy of a pair of nucleons as a function of their separation. Indicate the region in which nuclear force is a) attractive b) repulsive
ii) How proton and neutron exist together in an extremely small space within the nucleus whereas protons have force of repulsion between them?
Because of strong nuclear force which acts between p-p , p-n and n-n all it always attractive.
iii) Why heavy stable nuclei mostly contain lesser protons than neutrons?
Because protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so that an excess of neutrons which produce only attractive forces is required for stability
7. i) In Young’s double slit experiment, deduce the conditions for a) constructive and b) destructive interference at a point on the screen.
Conditions for constructive and destructive interference
Amplitude of electric field vectors are a1 and a2 respectively.
Wave equation is represented by,
Y1 = a1 sin ωt —- (i)
Y2 = a2 sin (ωt + Ø) —- (ii)
Using the theory of superposition,
Y = Y1 + Y2 —- (iii)
Here, Y1 and Y2 are the points of electric field.
Putting values from (ii) and (iii) in (i), we have
Y = a1 sin ωt + a2 sin (ωt + Ø)
Now using trigonometric identities, we have
sin (ωt + Ø) = sin ωt cos Ø + cos ωt sin Ø
we get,
Y = a1 sin ωt + a2 (sin ωt cos Ø + cos ωt sin Ø)
= (a1 + a2 cos Ø) sin ωt + (a2 sin Ø) cos ωt —- (iv)
Assume,
a1 + a2 cos Ø = A cos θ
And
a2 sin Ø = A sin θ
So, equ (iv) gives,
Y = A cos θ sin ωt + A sin θ cos ωt
= A sin (ωt + Ø)
Amplitude of the resultant wave is given by,
Amplitude, A = √(a12 + a22 + 2a1a2cos Ø)
Intensity of the wave is proportional to the amplitude of the wave.
Thus, intensity of the resultant wave is given by.
I = a12 + a22 + 2a1a2cos Ø
Constructive interference: for maximum intensity at any point, cos = + 1
Phase difference, Ø = 0, 2π, 4π, 6π, …..
= 2nπ (n = 0, 1, 2, ….)
So, maximum intensity is,
Imax = a12 + a22 + 2a1a2 = (a1 + a2)2
Path difference is,
Δ = λ/2π x Phase difference = λ/2π x 2nπ = nλ
Constructive interference is obtained when the path difference between the waves is an integral multiple of λ
Destructive interference
Ø = π, 3π, 5π, 7π, …..
= (2n – 1) π, n = 1, 2, 3, …..
Minimum intensity is,
Imin = a12 + a22 – 2a1a2 = (a1 – a2)2
Path difference is,
Δ = λ/2π x Phase difference
= λ/2π x (2n – 1)π = (2n – 1)λ/2
In destructive interference, path difference is odd multiple of λ/2
ii) What is maximum number of possible interference maxima for YDSE in which slit separation is twice the wavelength of monochromatic light?
Ans. For possible interference maxima on the screen, the condition is
d sin θ = nλ …. (i)
Given, d = slit-width = 2λ
Therefore, 2λ sin θ = nλ
Or 2 sin θ = n
The maximum value of sin θ is 1, hence
N = 2 x 1 = 2
Thus, Eq (i) must be satisfied by 5 integer values i.e., -2, -1, 0, 1, 2. Hence, the maximum number of possible interference maxima is 5.
8. A biconvex lens (n=3/2) has radii of curvature 20cm each. It is fitted into a hole in a large box filled with water (n= 4/3). A point object is placed outside the box at a distance of 40cm from the lens on its axis. Find the distance of the image formed in water in the box.
Ans. Since the material on both sides of the lens is different the twin lens formula cannot be used directly. We will therefore apply the spherical surface refraction formula for the two spherical surfaces formed on either side of the lens.
The rays of light diverging from the lens undergo the first refracting surface which is formed by air and glass. At this surface,
µ1 = 1, µ2 = 3/2, u = -40 cm
and 3/2v – 1/-40, R = 20 cm
or v = ∞
For the second surface now, the object distance will be infinite as the rays become parallel after the first refraction. In case of the second surface
µ1 = 3/2, µ2 = 4/3, u = -20 cm
Hence, 4/3v – 3/2(∞) = 4/3v = + 1/120
v = + 160 cm
Hence, a real image is formed in water.
OR
A ray PQ incident normally on the refracting face BA is refracted in the Prism BAC made of material of refractive index 1.5. Complete the path ray through the prism. From which face will the ray emerge? Justify your answer.
Ans. The incidence angle is given as,
sin ic = 2/3
sin ic = 0.66
So, ic > 300
Thus, the light will emerge out from face AC
9. Find the ratio of kinetic energy of the particle to the energy of the photon, if the de Broglie wavelength of a particle moving with a velocity 2.25×108 m/s is equal to the wavelength of photon.
10. i) Draw the labelled ray diagram for the formation of image by a compound microscope. Derive an expression for its total magnification, when the final image is formed at the near point.
(a) Image formation by a compound microscope: A schematic diagram of a compound microscope is shown in fig
(b) Magnifying Power: The linear magnification (m0) due to the objective is
When h’ is the size of the first image, the object size being h and f0 being the focal length of the objective and L be the distance between the second focal point of the objective and first focal point of the eye piece (focal length fe) is called the tube length of compound microscope.
When the final image is formed at the near point, then the angular magnification (me) of the eye piece is
ii) Why both objective and eyepiece of a compound microscope must have short focal lengths?
Ans. f0 and fe are in denominator.
This formula contains f0 and fe in denominator. Therefore, both the objective and the eyepiece of a compound microscope must have short focal lengths.
11. i) Draw a sketch of linearly polarized electromagnetic waves propagating in the Zdirection. Indicate the directions of the oscillating electric and magnetic fields
ii) Identify the electromagnetic waves whose wavelengths lie in the range.
a) 10-14m < λ < 10-11 m b) 10-6 m < λ < 10-4 m
Write one use of each
Ans. The range of wavelength of gamma rays is typically 10-11 m to 10-14 m and the wavelength range of infrared is 10-4 m to 10-6 m.
The gamma rays are used to treat some types of cancer, since the rays also kill cancer cells.
The infrared is used in heat sensitive thermal imaging cameras. It can be used to study human and animal body heat patterns.
OR
State Huygens’s principle. Using Huygens’s constructions draw a figure showing the propagation of a plane wave refracting at a plane surface separating two mediums. Hence verify Snell’s law of refraction.
Huygens’ Principle is based on the following assumptions
Each point on the primary wavefront acts as a source of secondary wavelets, sending out disturbances in all directions in a similar manner as the original source of light does.
SECTION – C
12. Total internal reflection, in physics, complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. The phenomenon occurs if the angle of incidence is greater than a certain limiting angle, called the critical angle. In general, total internal reflection takes place at the boundary between two transparent media when a ray of light in a medium of higher index of refraction approaches the other medium at an angle of incidence greater than the critical angle.
I) What is the relation between critical angle and refractive index?
a) μ = cosC b) μ = 1/cosC c) μ = 1/sinC d) μ = sinC
II) Critical angle for glass air interface where µ of glass is 3/2 is
(a) 41.8° (b) 60° (c) 30° (d) 44.3°
III) Critical angle for water air interface for violet colour is 49°. Its value for red colour would be
(a) 49° (b) 50° (c) 48° (d) 52°
IV) If the critical angle for total internal reflection from a medium to vacuum is 30°, then the velocity of light in the medium is,
(a) 3 × 108 m/s (b) 1.5 × 108 m/s (c) 6 × 108 m/s (d) 3 × 108 m/s
V) Critical angle of light passing from glass to water is minimum for
(a) red colour (b) green colour (c) yellow colour (d) violet colour