Ray Optics and Optical Instruments


Ray Optics and Optical Instruments

  1. The radius of curvature of a thin plano-convex lens is 10 cm (of curved surface) and the refractive index is 1.5. If the plane surface is silvered, then it behaves like a concave mirror of focal length​









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    The silvered plano convex lens behaves as a concave mirror ; whose focal length is given by

    1
    =
    2
    +
    1
    Ff1fm


    If plane surface is silvered
    fm =
    R2
    =
    = ∞
    22

    1
    = (μ - 1)
    1
    -
    1
    f1R1R2

    = (μ - 1)
    1
    -
    1
    =
    μ - 1
    RR

    1
    =
    2(μ - 1)
    +
    1
    =
    2(μ - 1)
    FRR

    F =
    R
    2(μ - 1)

    Here, R = 10 cm, µ = 1.5
    ∴ F =
    10
    = 10 cm
    2(1.5 - 1)

    Correct Option: A

    The silvered plano convex lens behaves as a concave mirror ; whose focal length is given by

    1
    =
    2
    +
    1
    Ff1fm


    If plane surface is silvered
    fm =
    R2
    =
    = ∞
    22

    1
    = (μ - 1)
    1
    -
    1
    f1R1R2

    = (μ - 1)
    1
    -
    1
    =
    μ - 1
    RR

    1
    =
    2(μ - 1)
    +
    1
    =
    2(μ - 1)
    FRR

    F =
    R
    2(μ - 1)

    Here, R = 10 cm, µ = 1.5
    ∴ F =
    10
    = 10 cm
    2(1.5 - 1)


  1. A body is located on a wall. Its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance 'd' ahead of second wall, then the required focal length will be​









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    Using the lens formula ,
    1
    =
    1
    -
    1
    fvu

    Given v = d, for equal size image ​| v | = | u | = d
    By sign convention u = –d
    1
    =
    1
    +
    1
    or f =
    d
    fdd2

    Correct Option: B

    Using the lens formula ,
    1
    =
    1
    -
    1
    fvu

    Given v = d, for equal size image ​| v | = | u | = d
    By sign convention u = –d
    1
    =
    1
    +
    1
    or f =
    d
    fdd2



  1. A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will​









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    1
    = (μg - 1)
    1
    -
    1
    fR1R2

    where , μg = 1 is given
    1
    = (1 - 1)
    1
    -
    1
    = 0 ⇒ f = ∞
    fR1R2

    Correct Option: C

    1
    = (μg - 1)
    1
    -
    1
    fR1R2

    where , μg = 1 is given
    1
    = (1 - 1)
    1
    -
    1
    = 0 ⇒ f = ∞
    fR1R2


  1. An equiconvex lens is cut into two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let f, f ', f '' be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively.​​​

    Choose the correct statement from the following ​









  1. View Hint View Answer Discuss in Forum


    1
    = (μ - 1)
    1
    -
    1
    fR1R2

    In this case, R1 and R2 are unchanged ​
    So, f will remain unchanged for both pieces of the lens
    ∴ f = f '​

    1
    =
    1
    +
    1
    ff1f2

    This is combination of two lenses of equal focal lengths
    1
    =
    1
    +
    1
    =
    2
    ⇒ f" = 2f
    ff"f"f"

    Correct Option: B


    1
    = (μ - 1)
    1
    -
    1
    fR1R2

    In this case, R1 and R2 are unchanged ​
    So, f will remain unchanged for both pieces of the lens
    ∴ f = f '​

    1
    =
    1
    +
    1
    ff1f2

    This is combination of two lenses of equal focal lengths
    1
    =
    1
    +
    1
    =
    2
    ⇒ f" = 2f
    ff"f"f"



  1. A convex  lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is​









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    From the formula,

    1
    =
    1
    +
    1
    =
    1
    -
    1
    = 0
    ff1f22525

    Power of combination =
    1
    = 0
    f

    Correct Option: C

    From the formula,

    1
    =
    1
    +
    1
    =
    1
    -
    1
    = 0
    ff1f22525

    Power of combination =
    1
    = 0
    f