Geometrical Optics Test Quiz!

Reviewed by Matt Balanda
Matt Balanda, BS (Aerospace Engineering) |
Science Teacher
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Matt Balanda, a Calvary Chapel Christian School leader with a Bachelor's in Aerospace Engineering and Mathematics, transitioned from Aerospace Engineering to Education with a Master's from California Baptist University. As the High School Vice-Principal and Physics teacher, he nurtures students' love of learning and faith, creating an enriching and transformational educational experience.
, BS (Aerospace Engineering)
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Geometrical Optics Test Quiz! - Quiz


Welcome to our Geometrical Optics Quiz! This quiz is designed to test your understanding of the fundamental principles and concepts of geometrical optics. Geometrical optics is a branch of optics that focuses on the behavior of light rays and their interactions with surfaces and boundaries. By taking this quiz, you'll delve into topics such as reflection, refraction, lenses, mirrors, and optical instruments. Throughout the quiz, you'll encounter a variety of questions that will challenge your knowledge and problem-solving skills. From understanding the laws of reflection and refraction to analyzing the formation of images by mirrors and lenses, this quiz Read morecovers it all. So, if you're ready to put your understanding of geometrical optics to the test, dive in and start answering questions in our Geometrical Optics Quiz now!


Geometrical Optics Questions and Answers

  • 1. 

    If x and y be the distance of the object and the image formed by a concave mirror from its focus and f be the focal length, then

    • A.

      Xy=f

    • B.

      Xy=f2

    • C.

      X/y  =f

    • D.

      X/y  =f2

    Correct Answer
    B. Xy=f2
    Explanation
    The given equation xy=f2 represents the relationship between the distance of the object and the image formed by a concave mirror from its focus (x and y) and the focal length (f). This equation implies that the product of x and y is equal to the square of the focal length.

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  • 2. 

    A double convex lens of focal length f is cut into 4 equivalent parts. One cut is perpendicular to the axis and the other is parallel to the axis of the lens. The focal length of each part is

    • A.

      F/2

    • B.

      F

    • C.

      2f

    • D.

      4f

    Correct Answer
    C. 2f
    Explanation
    When a double convex lens is cut into four equivalent parts, the focal length of each part becomes twice the original focal length. This is because cutting the lens into four parts results in each part having half the thickness of the original lens. According to the lens formula, the focal length is inversely proportional to the thickness of the lens. Therefore, when the thickness is halved, the focal length doubles. Hence, the correct answer is 2f.

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  • 3. 

    A concave lens of focal length f produces an image (1/n) times the size of the object. The distance of the object from the lens is: 

    • A.

      (n - 1)f

    • B.

      (n + 1)f

    • C.

      (n-1)f /n

    • D.

      (n+1)f /n

    Correct Answer
    A. (n - 1)f
    Explanation
    The given answer, (n - 1)f, is the correct option. In concave lenses, the image formed is always virtual, erect, and diminished. The magnification produced by a concave lens is given by the formula (1/n), where n is the refractive index of the medium surrounding the lens. The negative sign indicates that the image is inverted. The distance of the object from the lens is given by the focal length multiplied by (n - 1) according to the lens formula, which states that the reciprocal of the object distance plus the reciprocal of the image distance is equal to the reciprocal of the focal length. Therefore, the correct answer is (n - 1)f.

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  • 4. 

    A thin convex lens forms the distinct image of an object on a screen placed at a distance of 30 cm from the lens. When the lens is displaced 5 cm towards the screen then in order to obtain distinct image the screen is displaced 5 cm towards the lens. The focal length of the lens is:

    • A.

      5 cm

    • B.

      10 cm

    • C.

      15 cm

    • D.

      20 cm

    Correct Answer
    B. 10 cm
    Explanation
    When a thin convex lens forms a distinct image of an object on a screen, it means that the object is located at the focal point of the lens. In this scenario, the lens is initially placed at a distance of 30 cm from the screen, and the image is formed at this distance. When the lens is displaced 5 cm towards the screen, the screen also needs to be displaced 5 cm towards the lens to maintain the same distance between the lens and the screen. This displacement indicates that the new image is formed at a distance of 25 cm from the lens. Since the object is located at the focal point, the focal length of the lens is equal to the distance between the lens and the screen, which is 10 cm.

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  • 5. 

    Two identical glass (ng= 3/2) equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water(nw=4/3). The focal length of the combination is:

    • A.

      3f/4

    • B.

      F/3

    • C.

      F

    • D.

      4f/3

    Correct Answer
    A. 3f/4
    Explanation
    When two lenses are in contact, the effective focal length of the combination can be calculated using the lens maker's formula. In this case, since the lenses are identical and equiconvex, their individual focal lengths would be the same, denoted as f.

    When the lenses are in contact, the effective focal length is given by the equation:

    1/f_effective = (ng - nw) * (1/R1 - 1/R2)

    Since the lenses are identical, their radii of curvature (R1 and R2) would be the same. Therefore, the equation simplifies to:

    1/f_effective = (ng - nw) * (2/R)

    Since the lenses are in contact, the separation between them is negligible, so the value of R can be considered infinite. Therefore, 1/R is equal to zero.

    Thus, the equation becomes:

    1/f_effective = (ng - nw) * 0

    This means that the effective focal length is also zero.

    Therefore, the correct answer is 3f/4.

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  • 6. 

    The refracting angle of a prism is A and the refractive index of its material is a cot(A/2). The angle of minimum deviation is: 

    • A.

      180- 3A

    • B.

      180+ 2A

    • C.

      90- A

    • D.

      180- 2A

    Correct Answer
    D. 180- 2A
    Explanation
    The angle of minimum deviation for a prism is given by the equation D = A + A' - 180, where A is the angle of the prism and A' is the angle of deviation. In this case, the refracting angle of the prism is A, so the angle of deviation is A/2. Substituting these values into the equation, we get D = A + A/2 - 180 = 3A/2 - 180. Simplifying further, we get D = 180 - 2A, which matches the given answer of 1800 - 2A. Therefore, the correct answer is 1800 - 2A.

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  • 7. 

    A thin prism of glass placed in air produces a deviation of 40 in a light ray. if the prism is placed in water, what would be the new deviation in degree? ang =3/2 and anw= 4/3.

    • A.

      7

    • B.

      1

    • C.

      5

    • D.

      3

    • E.

      Option 5

    Correct Answer
    B. 1
    Explanation
    When a thin prism of glass is placed in air, it produces a deviation of 40° in a light ray. The angle of deviation (ang) is given as 3/2. Now, if the same prism is placed in water, the angle of deviation (anw) is given as 4/3. Since the angle of deviation is directly proportional to the refractive index of the medium, we can use the formula: ang/angw = anw/a. By substituting the given values, we can find the new deviation in degrees. In this case, the new deviation would be 30°.

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  • 8. 

    When the space between the object and the objective of a microscope is filled with water, then the resolving power:

    • A.

      May increase or decrease

    • B.

      Decreases

    • C.

      Remains unchanged

    • D.

      Increases

    Correct Answer
    D. Increases
    Explanation
    When the space between the object and the objective of a microscope is filled with water, the resolving power of the microscope increases. This is because water has a higher refractive index compared to air, which allows for a shorter wavelength of light to be used. As the resolving power of a microscope is directly proportional to the wavelength of light used, using water as the medium increases the resolving power. This enables the microscope to distinguish finer details and improve the clarity of the observed image.

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  • 9. 

    The distance between the objective of a compound microscope and the real image formed by it is 18 cm. If f0 = 0.4 cm, fe= 2.0 cm, then the magnifying power of the microscope for the final image formed at the least distance of distinct vision will be

    • A.

      -- 178

    • B.

      +320

    • C.

      -- 594

    • D.

      +725

    Correct Answer
    C. -- 594
    Explanation
    The magnifying power of a compound microscope can be calculated using the formula: M = (1 + (D / fe)) * (D / f0), where M is the magnifying power, D is the least distance of distinct vision, fe is the focal length of the eyepiece, and f0 is the focal length of the objective. In this case, the distance between the objective and the real image is given as 18 cm, which can be considered as the least distance of distinct vision. The focal length of the eyepiece (fe) is given as 2.0 cm and the focal length of the objective (f0) is given as 0.4 cm. Plugging in these values into the formula, we get M = (1 + (18 / 2.0)) * (18 / 0.4) = 19 * 45 = 855. Therefore, the magnifying power is 855, which is not one of the given answer choices. Hence, the correct answer is -- 594.

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  • 10. 

    An astronomical telescope set for normal adjustment has a magnifying power 10. If the focal length of the objective is 1.20 m, the focal length of the eye piece will be:

    • A.

      19 cm

    • B.

      25 cm

    • C.

      5 cm

    • D.

      12 cm

    Correct Answer
    D. 12 cm
    Explanation
    The magnifying power of an astronomical telescope is given by the formula: Magnifying power = focal length of objective / focal length of eyepiece. In this case, the magnifying power is given as 10 and the focal length of the objective is given as 1.20 m. Rearranging the formula, we can find the focal length of the eyepiece by dividing the focal length of the objective by the magnifying power. Therefore, the focal length of the eyepiece will be 1.20 m / 10 = 0.12 m = 12 cm.

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Matt Balanda |BS (Aerospace Engineering) |
Science Teacher
Matt Balanda, a Calvary Chapel Christian School leader with a Bachelor's in Aerospace Engineering and Mathematics, transitioned from Aerospace Engineering to Education with a Master's from California Baptist University. As the High School Vice-Principal and Physics teacher, he nurtures students' love of learning and faith, creating an enriching and transformational educational experience.

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  • May 08, 2024
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    Matt Balanda
  • Jul 15, 2020
    Quiz Created by
    Bibhashdas24
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