12th Grade Physics Exam: Quiz!
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An electron, an alpha-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?
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A. electron
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B. proton
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C. alpha-particle
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Preparing for the 12th-grade physics exam? Ready to take the quiz? As it pertains to this quiz, you must understand how many electron volts make up 1 joule and the electrons ejected from metals because of the effect of light. What is the refractive directory of the core in optical fibers, the angle of minimum deviation, and a wavefront See moredue to a source situated at infinity? Good luck with this quiz.
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- 2.
How many electron volts makeup 1 Joule?
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(a) 6.25×10^18 eV.
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(b) 1.25×10^181 eV
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(c) 6.25×10^16 eV.
Correct Answer
A. (a) 6.25×10^18 eV.Explanation
1 eV= 1.6 × 10 ^−19 J
So 1 J= 6.25×10 ^18 eV
Hence (a) is the correct optionRate this question:
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- 3.
If the maximum Kinetic energy of electrons emitted in a photo-cell is 6 eV. What is the stopping potential?
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3 V
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12 V
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6 V
Correct Answer
A. 6 VExplanation
The stopping potential in a photo-cell is the minimum potential difference required to stop the flow of electrons. In this case, the maximum kinetic energy of the emitted electrons is given as 6 eV. The stopping potential is equal to the maximum kinetic energy, which means it is also 6 V.Rate this question:
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- 4.
The electrons ejected from metals due to effect of light are called
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Primary electrons
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Thermions
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Photoelectrons
Correct Answer
A. PhotoelectronsExplanation
When light shines on a metal surface, it can cause the electrons in the metal to be ejected. These ejected electrons are called photoelectrons. This phenomenon is known as the photoelectric effect. The energy of the incident light is transferred to the electrons, giving them enough energy to overcome the attractive forces within the metal and escape. These photoelectrons can be detected and measured, providing valuable information about the properties of the metal and the characteristics of the incident light.Rate this question:
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- 5.
The frequency of the incident light falling on a photosensitive metal plate is doubled, stopping potential will be:
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Double the earlier value
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Unchanged
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More than doubled
Correct Answer
A. More than doubledExplanation
ⅇV0=hν-∅0 use this formula. ⅇV0 + ∅0 =hνRate this question:
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- 6.
A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen.
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(a) half the image will disappear.
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(b) intensity of image will decrease but complete image is formed.
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(c) intensity of image will increase but image is not distinct.
Correct Answer
A. (b) intensity of image will decrease but complete image is formed.Explanation
because focal length of lens does not change but amount of light passing through lens becomes half.Rate this question:
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- 7.
In optical fibers, the refractive index of the core is:
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Greater than that of the cladding.
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Smaller than that of the cladding.
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Equal to that of the cladding.
Correct Answer
A. Greater than that of the cladding.Explanation
R.I. of core is greater than that of the cladding for total internal reflection to occur.Rate this question:
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- 8.
A metal coin is at bottom of a beaker filled with a liquid of refractive index = 4/3 to a height of 6 cm. To an observer looking from above the surface of the liquid, coin will appear at a depth.
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(a) 1.5 cm
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(b) 4.5 cm
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(c) 6.75 cm
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Option 4
Correct Answer
A. (b) 4.5 cmExplanation
R.I = Real depth / apparent depthRate this question:
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- 9.
The refractive index of the material of an equilateral prism is √3. What is the angle of minimum deviation?
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(a) 45°
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(b) 60°
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(c) 37°
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Option 4
Correct Answer
A. (b) 60°Explanation
use Prism formula for minimum deviation conditionRate this question:
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- 10.
In an experiment to find the focal length of a concave mirror, a graph is drawn between the magnitude of u and v. The graph looks like?
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Option 1
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Option 2
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Option 3
Correct Answer
A. Option 3Explanation
u – v curve is a rectangular parabola.Rate this question:
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- 11.
The wavefront due to a source situated at infinity is:
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Spherical
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Planar
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Cylindrical
Correct Answer
A. PlanarExplanation
When a source is situated at infinity, the wavefront it produces appears as a planar wavefront. This means that the wavefront is flat and extends infinitely in all directions. A planar wavefront is characterized by parallel wavefront lines, which indicates that the wave is propagating in a straight line. In contrast, a spherical wavefront would indicate that the wave is originating from a point source, and a cylindrical wavefront would suggest that the wave is spreading out in a cylindrical shape. Therefore, the correct answer is planar.Rate this question:
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- 12.
A laser beam is coherent because it contains:
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(a) waves of several wavelengths.
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(b) coherent waves of a single wavelength.
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(c) coherent waves of several wavelengths
Correct Answer
A. (b) coherent waves of a single wavelength.Explanation
A laser beam is coherent because it contains coherent waves of a single wavelength. Coherence refers to the property of waves having a constant phase relationship with each other. In a laser, the light waves are produced through stimulated emission, where photons are emitted in phase with each other. This results in a beam of light with a single wavelength and a constant phase relationship between the waves. This coherence allows the laser beam to have a narrow, focused beam and to exhibit interference effects such as diffraction.Rate this question:
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- 13.
According to Huygens’ principle, light is a form of
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Particle
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Wave
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Radiation
Correct Answer
A. WaveExplanation
According to Huygens' principle, light is considered to be a wave. This principle states that every point on a wavefront acts as a source of secondary spherical waves, and the sum of these waves determines the new position of the wavefront. This concept is used to explain various optical phenomena, such as diffraction and interference, which are characteristic of wave behavior. Therefore, the correct answer is wave.Rate this question:
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- 14.
Two coherent monochromatic light beams of intensities I and 41 superimpose. The maximum and minimum possible intensities in the resulting beam are:
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(a) 5I and I
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(b) 5I and 3I
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(d) 9I and I
Correct Answer
A. (d) 9I and IExplanation
I max= I1 + I2 + 2 √ (I1 X I2)= 9 I
I min = I1 + I2 - 2 √ (I1+I2) = 1 IRate this question:
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Nov 04, 2024Quiz Edited by
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Aug 14, 2020Quiz Created by
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