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2.4: Convergence

  • Page ID
    7076
  • [ "article:topic", "authorname:tatumj", "Convergence (Optics)" ]

    Figure II.5 shows a lens made of glass of refractive index 1.50. To the left of the lens is air (refractive index 1.00). To the right of the lens is water (refractive index 1.33). A converging beam of light is incident upon the lens directed toward a virtual object O that is 60 cm from the lens. After refraction through the lens, the light converges to a real image I that is 20 cm from the lens.

    I am not at this stage going to ask you to calculate the radii of curvature of the lens. (You can’t – you need one more item of information.) I just want to use this diagram to define what I mean by convergence.

    The convergence of the light at the moment when it is incident upon the lens is called the initial convergence \(C_1\), and it is defined as follows:

    \[ initial\space convergence = \frac{Refractive\space index}{Object \space distance}. \label{eq:2.4.1} \]

    The convergence of the light at the moment when it leaves the lens is called the final convergence \(C_2\), and it is defined as follows:

    \[ final\space convergence = \frac{Refractive\space index}{Image \space distance}. \label{eq:2.4.2} \]

    Sign convention

    • Converging light has positive convergence;
    • Diverging light has negative convergence.

    Example \(\PageIndex{1}\)

    Initial convergence = \(+ \frac{1.00}{60}=+0.01667\) cm-1.

    Final convergence = \(+ \frac{1.33}{20}=+0.06650\) cm-1.

    Notice that, before the light enters the lens, it is in a medium of refractive index 1.00. Thus the relevant refractive index is 1.00, even though the virtual object is in the water.

    Contributor