2.4: Convergence
- Page ID
- 7076
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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:
\[ initial\space convergence = \frac{Refractive\space index}{Object \, distance}. \label{eq:2.4.1} \]
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:
\[ final\space convergence = \frac{Refractive\space index}{Image \, distance}. \label{eq:2.4.2} \]
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.