Physical Optics Quiz! Trivia

Reviewed by Matt Balanda
Matt Balanda, BS (Aerospace Engineering) |
Physics
Review Board Member
Matt holds a Bachelor's of Science in Aerospace Engineering and Mathematics from the University of Arizona, along with a Master's in Educational Leadership for Faith-Based Schools from California Baptist University. A devoted leader, he transitioned from Aerospace Engineering to inspire students. As the High School Vice-Principal and a skilled Physics teacher at Calvary Chapel Christian School, his passion is nurturing a love for learning and deepening students' connection with God, fostering a transformative educational journey.
, BS (Aerospace Engineering)
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Physical Optics Quiz! Trivia - Quiz

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Welcome to our Physical Optics Quiz! Dive into the fascinating world of optics with this engaging quiz that covers a wide range of topics related to physical optics. In this quiz, you'll explore interference, diffraction, polarization, and more concepts. Test your understanding of how light behaves and interacts with various mediums and structures. Prepare to enhance your knowledge and deepen your understanding of the fundamental principles of physical optics. Whether you're aiming to ace your optics exam or simply expand your knowledge of the natural world, this quiz is the perfect way to test your skills and Read morelearn something new along the way. So, are you ready to shine bright and tackle the challenges of our Physical Optics Quiz? Let's get started!


Physical Optics Questions and Answers

  • 1. 

    A beam of electrons is used in Young's double-slit experiment. If the speed of electrons is increased then the fringe width will:

    • A.

      Increase

    • B.

      Decrease

    • C.

      Remains same

    • D.

      Fringes will not be seen

    Correct Answer
    B. Decrease
    Explanation
    In Young's double-slit experiment, the fringe width is determined by the wavelength of the electrons. According to the equation for fringe width, w = λD/d, where w is the fringe width, λ is the wavelength, D is the distance from the slits to the screen, and d is the distance between the slits. If the speed of the electrons is increased, their wavelength will decrease due to the de Broglie equation, λ = h/p, where h is Planck's constant and p is the momentum of the electrons. As a result, the fringe width will decrease.

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  • 2. 

    In YDSE, the slits are 0.05 cm apart and the interference fringes are obtained on a screen 1 m away from the slits. The slits are illuminated by sodium light(5893 A). The distance between 4th bright fringe on one side and 3rd bright fringe on the other side of the central bright fringe is:

    • A.

      1.25 mm

    • B.

      5.5 mm

    • C.

      8.25 mm

    • D.

      3.78 mm

    Correct Answer
    C. 8.25 mm
    Explanation
    The distance between the 4th bright fringe on one side and the 3rd bright fringe on the other side of the central bright fringe can be calculated using the formula for fringe separation in Young's double-slit experiment:

    Δy = (λD) / d

    Where Δy is the fringe separation, λ is the wavelength of light, D is the distance between the screen and the slits, and d is the distance between the slits.

    Given that the slits are 0.05 cm apart (0.0005 m), the screen is 1 m away from the slits, and the wavelength of sodium light is 5893 A (589.3 nm or 0.0005893 m), we can calculate the fringe separation.

    Δy = (0.0005893 m * 1 m) / 0.0005 m = 1.1786 m

    Since the question asks for the distance in millimeters, we convert the result:

    1.1786 m * 1000 = 1178.6 mm

    The distance between the 4th bright fringe on one side and the 3rd bright fringe on the other side of the central bright fringe is approximately 1178.6 mm.

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  • 3. 

    Two slits in YDSE have widths in the ratio 1:25. The ratio of intensity at the maxima and minima in the interference pattern, Imax / Imin:

    • A.

      121/49

    • B.

      49/121

    • C.

      4/9

    • D.

      9/4

    Correct Answer
    D. 9/4
    Explanation
    In Young's double-slit experiment, the intensity of light at the maxima and minima in the interference pattern is determined by the square of the amplitude of the resultant wave at that point. The amplitude of the resultant wave is determined by the superposition of waves from both slits. Since the widths of the slits are in the ratio 1:25, the amplitudes of the waves from the slits will also be in the same ratio. The intensity is proportional to the square of the amplitude, so the ratio of intensities at the maxima and minima will be (1/25)^2 : 1^2 = 1/625 : 1 = 1 : 625. Simplifying this ratio gives 9/4, which is the ratio of Imax/Imin.

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  • 4. 

    The formula for the width of the central maximum in the Fraunhofer diffraction pattern of a single slit as observed on a screen:

    • A.

      F λ/e

    • B.

      2λ/e

    • C.

      F λ/2e

    • D.

      2f λ/e

    Correct Answer
    D. 2f λ/e
    Explanation
    The formula for the width of the central maximum in the Fraunhofer diffraction pattern of a single slit as observed on a screen is given by 2f λ/e. This formula takes into account the focal length (f) of the lens, the wavelength of light (λ), and the width of the slit (e). It suggests that the width of the central maximum is directly proportional to the product of the focal length and the wavelength, and inversely proportional to the width of the slit. This means that as the focal length or wavelength increases, or as the width of the slit decreases, the width of the central maximum will also increase.

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  • 5. 

    The angular separation between central maximum and first-order maximum of the diffraction pattern due to a single slit of width 0.25 mm when light of wavelength 5890 A is incident normally on it, is:

    • A.

      35 rad

    • B.

      3.5 rad

    • C.

      0.035 rad

    • D.

      0.0035 rad

    Correct Answer
    D. 0.0035 rad
    Explanation
    When light passes through a single slit, it diffracts and forms a diffraction pattern. The central maximum is the bright spot in the center of the pattern, and the first-order maximum is the first bright spot on either side of the central maximum. The angular separation between these two maxima can be calculated using the formula: θ = λ / w, where θ is the angular separation, λ is the wavelength of light, and w is the width of the slit. Plugging in the values given in the question (λ = 5890 A = 5890 x 10^-10 m and w = 0.25 mm = 0.25 x 10^-3 m), the angular separation is calculated to be 0.0035 rad.

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  • 6. 

    In the propagation of light waves, the angle between the direction of propagation and the plane of polarisation is:

    • A.

      0

    • B.

      450

    • C.

      900

    • D.

      1800

    Correct Answer
    A. 0
    Explanation
    The angle between the direction of propagation and the plane of polarization is 0 degrees. This means that the light waves are propagating in the same plane as the polarization of the light. In other words, the electric field of the light waves is oscillating in a single plane as it travels.

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  • 7. 

    A beam of unpolarised light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that its principal plane makes an angle of 45relative to that of A. The intensity of the emergent light is: 

    • A.

      I0

    • B.

      I0 /2

    • C.

      I0 /4

    • D.

      I0 /8

    Correct Answer
    C. I0 /4
    Explanation
    When unpolarized light passes through a polaroid, it becomes polarized in a single plane. When this polarized light passes through another polaroid, the intensity of the emergent light depends on the angle between the principal planes of the two polaroids. In this case, the angle between the principal planes of polaroid A and polaroid B is 45 degrees. When the angle between the principal planes is 45 degrees, the intensity of the emergent light is half of the original intensity. Therefore, the correct answer is I0/4.

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  • 8. 

    Monochromatic light of wavelength 198 nm is incident on the surface of a metallic cathode, whose work function is 2.5 eV. How much potential difference must be applied between the cathode and the anode of a photocell to just stop the photocurrent from flowing?

    • A.

      2.55 V

    • B.

      3.75V

    • C.

      5.75 V

    • D.

      1.5 V

    Correct Answer
    B. 3.75V
    Explanation
    When monochromatic light of a specific wavelength is incident on a metallic cathode, electrons are emitted from the cathode due to the photoelectric effect. To stop the photocurrent from flowing, a potential difference must be applied between the cathode and the anode to counteract the emission of electrons. The minimum potential difference required to stop the photocurrent is equal to the work function of the cathode. In this case, the work function is given as 2.5 eV, so the potential difference needed is 2.5 V. However, the question asks for the potential difference in volts, not electron volts. Since 1 eV is equal to 1.6 x 10^-19 J, the potential difference in volts is calculated by dividing the work function by the charge of an electron (1.6 x 10^-19 C). Thus, the potential difference required is 2.5 V / (1.6 x 10^-19 C) = 3.75 V. Therefore, the correct answer is 3.75 V.

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  • 9. 

    The work function of a surface of photo sensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5V lies in the?

    • A.

      Ultraviolet region

    • B.

      Visible region

    • C.

      Infrared region

    • D.

      X- ray region

    Correct Answer
    A. Ultraviolet region
    Explanation
    The work function of a material is the minimum amount of energy required to remove an electron from its surface. In this question, the stopping potential is given as 5V, which is the maximum kinetic energy that an electron can have after being emitted from the material. The stopping potential is related to the wavelength of the incident radiation by the equation E = hc/λ, where E is the energy of a photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of the radiation. By rearranging the equation, we can find that λ = hc/E. Plugging in the given values, we get λ = (6.63 x 10^-34 J.s x 3 x 10^8 m/s) / (5 eV x 1.6 x 10^-19 J/eV) = 2.48 x 10^-7 m. This wavelength corresponds to the ultraviolet region of the electromagnetic spectrum.

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  • 10. 

    The kinetic energy of a proton is equal to the energy E of a photon. If f1 be the de Broglie wavelength of the proton and f2 that of the photon, then the ratio f1/ f is proportional to:

    • A.

      E

    • B.

      E1/2

    • C.

      --1/2

    • D.

      --1

    Correct Answer
    B. E1/2
    Explanation
    The de Broglie wavelength of a particle is inversely proportional to its momentum. Since the kinetic energy of a particle is directly proportional to its momentum, we can say that the de Broglie wavelength is inversely proportional to the square root of the kinetic energy. Therefore, the ratio f1/f2 will be proportional to the square root of the ratio of the kinetic energies, which is E1/2.

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Matt Balanda |BS (Aerospace Engineering) |
Physics
Matt holds a Bachelor's of Science in Aerospace Engineering and Mathematics from the University of Arizona, along with a Master's in Educational Leadership for Faith-Based Schools from California Baptist University. A devoted leader, he transitioned from Aerospace Engineering to inspire students. As the High School Vice-Principal and a skilled Physics teacher at Calvary Chapel Christian School, his passion is nurturing a love for learning and deepening students' connection with God, fostering a transformative educational journey.

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  • May 08, 2024
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  • Jul 28, 2020
    Quiz Created by
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