UMKULTHOOM SCHOOL, AL AIN
GRADE 12 G
PHYSICS- EOT QUESTIONS TERM-3(2022-2023)
TEACHER-SHIKHA AWASTHI
Q1. Differentiate between incoherent and coherent waves by giving examples like rain or water
droplets falling on the surface of water.
(Page-184)
Ans- Incoherent light -It is light whose waves are not in phase. For example- white light
Coherent light- It is Light made up of waves of the same wavelength that are in phase with each
other. For example-laser light.
Q2. Explain how bright and dark interference fringes are created in a double-slit interference
investigation with monochromatic light.
(Page187)
ANS-The wavefronts interfere constructively and destructively to form a pattern of bright and dark
bands.
The diagram in the figure shows, that light that reaches point P0 travels the same distance from each
slit. Because the waves are in phase, they interfere constructively on the screen to create the central
bright band at P0.
There is also constructive interference at the first bright band (P1) on either side of the central band
because line segment P1S1 is one wavelength (λ) longer than the line segment P1S2. Thus, the waves
arrive at P1 in phase.
Q3. Explain the phenomenon of thin-film interference.
(Page190)
Ans-The interference occurs between two reflected waves, one reflects from the outer surface of the
thin film and the other from the inner surface of the thin film, and we can get two types of interference
(constructive or destructive)
Condition for inversion of reflected rayIf n2> n1, the reflected ray will invert.
If n2< n1, the reflected ray will invert.
For a film of varying thickness, the wavelength requirement will be met at different thicknesses for
different colors. The result is a rainbow of color.
Where the film is too thin to produce constructive interference for any wavelength of visible light, the
film appears to be black.
Q4. Define and interpret the emission spectrum of an element or an object. (Page211)
ANS-A graph of the intensity of the radiation emitted from an object over a range of frequencies is
known as an emission spectrum.
1- The radiation intensity increases when the temperature increases but it reaches a maximum value
and then start decreasing.
2- The frequency at which the maximum amount of energy is emitted increases when the temperature
increases.
3- The types of radiations that emitted from the object depend on its temperature.
The power (energy emitted per second) of an electromagnetic wave is proportional to the hot object’s
kelvin temperature raised to the fourth power T4
Q5. Explain that electrons are ejected from the surface of a metal only if the frequency of the
incident radiation is greater than a threshold frequency (f)which is characteristic of the metal.
(Page219)
ANS-1. Electrons eject from the cathode only if the frequency of the incident radiation is greater
than a certain minimum value, called the threshold frequency (f0)
2: When a photon of frequency f0 is incident on a metal, the energy of the photon is enough to release
an electron but not enough to provide the electron with any kinetic energy.
3.If the frequency of photon is less than f0(threshold frequency), no electrons will be emitted from the
metal surface.
The emission of electrons when electromagnetic radiation falls
on an object is called the photoelectric effect.
Q6. Describe and calculate de Broglie wavelength for a moving particle. (Page 223,229)
ANS- The De Broglie wavelength of a moving particle is equal to Planck’s constant divided by the
particle’s momentum.
ℎ
ℎ
λ = 𝑃 = 𝑚𝑣
Q7. Define a monochromatic light.
(Page 185)
Ans- Monochromatic light is light of only one wavelength that is (Single color)
Q8. Recall the concepts of constructive and destructive interference and define interference
fringes of light.
(Page 186)
Constructive- When two waves meet in such a way that their crests line up together. The phase
difference is zero or 2nπ.
In the interference pattern: The Bright band represents an area where a constructive interference
occurs.
Destructive- When two waves meet in such a way that the crest of one wave meets the trough of
another. The phase difference is π.
The dark area represents an area where a destructive interference occurs.
The pattern of bright and dark band is called interference fringes.
Q9. Compare the bright and dark bands from Young's Double Slit investigation with the
diffraction pattern from Single Slit Diffraction regarding the band spacing, light sources, width
of bands, and intensity of created bands. (Page 193)
Ans-When coherent, green light passes through a single, small opening that is between about 10 and
100 light wavelengths, the light is diffracted by both edges, and a series of bright and dark bands
appears on a distant screen, as shown in Figure.
Comparison - Instead of the nearly equally spaced bands produced by two coherent sources in
Young’s double-slit investigation, diffraction pattern has a wide, bright central band with dimmer,
narrower bands on either side.
When using red light instead of green, the width of the bright central band increases. With white light,
the pattern is a combination of patterns of all the colors of the spectrum.
Q10. Apply the equation for the width of the central bright band produced by a single slit
diffraction ( 2x1=2λ/W), where (2x1 ) is the width of the central bright band, ( λ) is the
wavelength, (L) is the distance to the screen, and (w) is the width of the slit. (Page 196,205)
16.Monochromatic green light of wavelength 546 nm falls on a single slit with a width of 0.095 mm.
The slit is located 75 cm from a screen. How wide will the central bright band be? (Ans- Xmin= 4.3mm)
17.Yellow light with a wavelength of 589 nm passes through a slit of width 0.110 mm and makes a
pattern on a screen. If the width of the central bright band is 2.60×10−2 m, how far is it from the slits to
the screen? (Ans- L=2.43m)
18.Light from a He-Ne laser (λ= 632.8 nm) falls on a slit of unknown width. A pattern is formed on a
screen 1.15 m away, on which the central bright band is 15.0 mm wide. How wide is the slit? (Answ=9.7x10-5m)
19.Yellow light falls on a single slit 0.0295 mm wide. On a screen that is 60.0 cm away, the central
bright band is 24.0 mm wide. What is the wavelength of the light? (Ans- λ=5.9x10 -7m)
20. White light falls on a single slit that is 0.050 mm wide. A screen is placed 1.00 m away. A student
first puts a blue-violet filter (λ= 441 nm) over the slit, then a red filter (λ= 622 nm). The student
measures the width of the central bright band.
a. Which filter produced the wider band?
Ans-Red, because central peak width is proportional to wavelength.
b. Calculate the width of the central bright band for both filters. (Ans- for Blue-2X1=18mm , For Red
2X1=25mm)
47.Monochromatic light passes through a single slit 0.010 cm wide and falls on a screen 100 cm
away, as shown in Figure 25. If the width of the central band is 1.20 cm, what is the wavelength of the
light? (ans-λ=6x10-7m)
Q11. Discuss the production and use of diffraction grating. Give examples on the applications
of diffraction gratings like in spectroscopes used for gemstone analysis or others. (Page
196,198)
Ans-A diffraction grating is a device that is made up of many small slits that diffract light and form a
pattern that is an overlap of single-slit diffraction patterns.
Diffraction gratings can have as many as 10,000 slits per centimeter, which means the spacing
between the slits can be as small as 10−6 m.
A diffraction grating is a useful tool for the study of light and objects that emit or absorb light. One
type of diffraction grating is called a transmission grating.
A transmission grating can be made by scratching very fine lines with a diamond point on glass that
transmits light. The spaces between the scratched lines act like slits.
Diffraction gratings can be used to enhance the appearance of diamonds. The gratings are etched
into certain surfaces of the diamond to improve the dispersion of light and make the gems appear
more brilliant.
An instrument used to measure light wavelengths using a diffraction grating is called a grating
spectroscope, as shown in the diagram in Figure. The source to be analyzed emits light that is
directed through a slit and a collimator and then to a diffraction grating. The grating produces a
diffraction pattern that is viewed through a telescope.
Diffraction gratings are incorporated into spectroscopes used to analyze gemstones. Experienced
gemologists recognize the patterns of bands produced by white light passing through different stones.
For example, three bright bands of green, yellow, and orange are a strong indication that cobalt is
present. This likely means that a blue stone is not an expensive gem such as sapphire or topaz, but
rather a cheap piece of glass that has been tinted blue.
Q12. Explain diffraction through circular apertures and discuss resolving of images using the
Rayleigh criterion.
(Page 200)
Ans-The circular lens of a telescope, a microscope, and even your eye acts as a hole, called an
aperture, through which light passes. An aperture diffracts light, just as a single slit does. Alternating
bright and dark rings occur with a circular aperture,
Rayleigh criterion states that if the center of the bright spot of one source’s image falls on the first
dark ring of the second , the two images are at the limit of resolution.
RAYLEIGH CRITERION The separation distance between objects that are at the limit of resolution is
equal to 1.22, times the wavelength of light, times the distance from the circular aperture to the
objects, divided by the diameter of the circular aperture.
Q13. Calculate the energy emitted or absorbed by a vibrating atom using the equation ( E=nhf
) where n is an integer, h is Planck's constant (h = 6.63×10-34 J/Hz ), and f is the frequency of
vibration. (Page 212)
Ans- ENERGY OF VIBRATION The energy emitted or absorbed by a vibrating atom is equal to the
product of an integer, Planck’s constant, and the frequency of vibration.
E=n hf
(n=0,1,2,3……..)
Let us try…..
Q- A photon has a frequency of 7.24×1 0 15 Hz. What is the energy of the photon?
E=4.78X10-18J)
ANS-
(Ans-
Q-The energy of a photon is determined by Planck's constant and the _____ of the photon.
a. size
b. quantum number
c. frequency
d. intensity
Q-What will be the frequency of a photon of energy 1.32 x10−10J? (Ans-f=2.0x1023Hz)
ANS-
Q14. Interpret the graph of maximum kinetic energy of an electron ejected from a metal versus
the frequency of incident photon, and determine graphically the threshold frequency and
Planck's constant.
(Page 219)
1. The slope equals the rise/run ratio of the line, which is equal to Planck’s constant (h).
2. In the graph-the threshold frequency (f0 ) is the point at which KE= 0.
Let us try……….
Q. From the graph -Find threshold frequency and Planck’s constant for sodium metal.
(Ans-f0=5.7x1014Hz, h=6.6x10-34J/Hz)
Q. From the graph -Find threshold frequency and Planck’s constant for magnesium metal.
(Ans-f0=8.84x1014Hz, h=6.6x10-34J/Hz)
Q15. Define the work function of a metal, and calculate its value. (Page 219)
WORKFUNCTION- A photon with that frequency has just enough energy to eject an electron from the
metal. This minimum energy is called the work function of a metal.
The magnitude of the work function is equal to h f 0.
When a photon of frequency f 0 is incident on a metal, the energy of the photon is sufficient to release
an electron but not sufficient to provide the electron with any kinetic energy.
LET US TRY………………….
EXAMPLE- A particular photocell uses a sodium cathode. Sodium has a threshold wavelength of 526
nm.
a. Find the work function of sodium in eV. (Ans=2.36 eV)
EXAMPLE2- The threshold wavelength of zinc is 310 nm. Find the threshold frequency, in Hz, and
the work function, in eV, of zinc. (Ans-f 0=9.7x1014Hz, W= 4eV)
Q16. Describe that collisions between photons and particle obey the laws of conservation of
energy and momentum. (Page 222)
ANS- The increase in wavelength that Compton observed meant that the X-ray photons had lost both
energy and momentum. The shift in the energy of scattered photons is called the Compton effect.
This shift in energy, indicated by a shift in wavelength, is very small—only about 10 -3 nm. It is a
measurable effect only when electromagnetic waves have wavelengths less than 10 -2 nm
Compton found that the energy and momentum gained by the ejected electrons equaled the energy
and momentum lost by the photons. Thus, photons obey the laws of conservation of energy and
momentum when they collide with other particles.
Q17. Apply the relation of the wavelength from double-slit investigation ( λ=xd/L) where 'x' is
the distance on the screen from the central bright fringe to the first bright band, 'd' is the
distance between the slits, and 'L ' is the distance from the slits to the screen. (Page
188,192,204)
1.Violet light falls on two slits separated by 1.90×10−5 m. A first-order bright band appears 13.2 mm
from the central bright band on a screen 0.600 m from the slits. What is λ? (Ans-λ=4.18x10-7m)
2.Yellow-orange light from a sodium lamp of wavelength 596 nm is aimed at two slits that are
separated by 1.90×10−5 m. What is the distance from the central band to the first-order yellow band if
the screen is 0.600 m from the slits?
(Ans- X 1 =1.88x10 -2 m)
3.In a double-slit investigation, physics students use a laser with λ= 632.8 nm. A student places the
screen 1.000 m from the slits and finds the first-order bright band 65.5 mm from the central line. What
is the slit separation?
(Ans-d= 9.66x10-6m)
4.Yellow-orange light with a wavelength of 596 nm passes through two slits that are separated by
2.25×10−5 m and makes an interference pattern on a screen. If the distance from the central line to
the first-order yellow band is 2.00×10−2 m, how far is the screen from the slits?
(Ans- L= 0.755 m)
14.Light of wavelength 542 nm falls on a double slit. Use the values from Figure to determine how far
apart the slits are?
(Ans-d=1.6x10-5m)
5.Light falls on a pair of slits that are 19.0 μm apart and located 80.0 cm from a screen, as shown in
Figure 23.The first-order bright band is 1.90 cm from the central bright band. What is the wavelength
of the light?
(Ans-λ=4.51x10-7 m)
Q18. Explain that constructive interference from a diffraction grating occurs at angles on
either side of the central bright line given by the equation m 𝜆 = dsinθ where m=1,2,3. (Page
199,206)
22.If blue light of wavelength 434 nm shines on a diffraction grating and the spacing of the resulting
lines on a screen that is 1.05 m away is 0.55 m, what is the spacing between the slits in the grating?
(Ans-d=9.4x10-7m)
23.A diffraction grating with slits separated by 8.60×10−7 m is illuminated by violet light with a
wavelength of 421 nm. If the screen is 80.0 cm from the grating, what is the separation of the lines in
the diffraction pattern?
(Ans-X=0.449m)
24.Blue light shines on the DVD in Example Problem 3. If the dots produced on a wall that is 0.65 m
away are separated by 58.0 cm, what is the wavelength of the light?
(Ans-λ=490nm)
25. Light of wavelength 632 nm passes through a diffraction grating and creates a pattern on a
screen that is 0.55 m away. If the first bright band is 5.6 cm from the central bright band, how many
slits per centimeter does the grating have?
(Ans-1.6x10-3slits/cm)
Q19. Calculate the kinetic energy of an electron ejected due to the photoelectric effect
KE=hf-hfo.
(Page 217,218,230)
Q- The stopping potential difference of a certain photocell is 4.0 V. How much kinetic energy does the
incident light give the electrons? Give your answer in both joules and electron volts.
(Ans-6.4x10-19 J,4 eV)
5. An electron has an energy of 2.3 eV. What is the kinetic energy of the electron in joules?
(Ans-E=3.7x10-19J)
6. What is the velocity of the electron in the previous problem?
(ANS-v=9.0x105 m/s)
7. What is the kinetic energy in eV of an electron with a velocity of 6.2×1 0 6 m/s? (Ans-v=1.1x102 eV)
8. The stopping potential for a photoelectric cell is 5.7 V. Calculate the maximum kinetic energy of the
emitted photoelectrons in eV.
Hint-KE=-qV0
(Ans-KE=5.7eV)
9. The stopping potential for a photoelectric cell is 5.1 V. How much kinetic energy does the incident
light give the electrons in joules? Hint-KE=-qV0
(Ans-KE=8.16x10-19J )
10. The maximum kinetic energy of emitted photoelectrons in a photoelectric cell is7.5×10 –19 J. What
is the stopping potential?
(Ans-V0=4.7V)
11.CHALLENGE The stopping potential required to prevent current through a photocell is 3.2 V.
Calculate the maximum kinetic energy in joules of the photoelectrons as they are emitted.
(Ans-K.E=5.12X10-19J)
70. Incident radiation falls on tin, as shown in Figure 16.The threshold frequency of tin is 1.2×10 15 Hz.
a. What is the threshold wavelength of tin? (Ans-λ=2.5x10-7 m)
b. What is the work function of tin? (Ans-w=8.0 x10-19 J)
c. The incident electromagnetic radiation has the wavelength indicated in Figure 16. What is the
kinetic energy of the ejected electrons in eV? (Ans-KE=2.4eV)
Q20. Define a photon, and calculate its energy. Calculate the momentum of a photon . (Page
228,231)
Ans- According to Einstein, visible light and other forms of electromagnetic radiation consist of
discrete, quantized bundles of energy known as photon.
The energy of a photon is equal to the constant 1240 eV·nm divided by the wavelength of the photon.
(1ev=1.6x10-19 J)
43. What is the momentum of a photon of violet light that has a wavelength of 4.0×10 2 nm?
(Ans-P=1.65x10-27kgm/s)
Q5.A photon has a frequency of 1.14×10
15
Hz. What is the energy of the photon? (HINT-E=hf)
A. 5.82×1 0 −49 J
C. 8.77×1 0 −19 J
B. 7.55×1 0 −19 J
D. 1.09×1 0 −12 J
Shikha Awasthi