GEOMETRICAL OPTICS: Positive and Negative Sign Conventions

In order for the Thin-Lens Equation and similar derivations to mathematically represent thin-lens systems, sign conventions must be assigned to the physical parameters encountered within optical systems that are modeled. In some circumstances, relationships between the height of an object ( h_o ) or image ( h_i ) are related to the distances ( d_o and d_i ) at which objects and images are situated relative to a lens. These relationships enable us to determine to what extent an image will be magnified:

M = ( h_i / h_o ) = ( – d_i / d_o )

As far as magnification ( M ) is concerned, a negative sign tells us that an image is inverted, whereas a positive sign lets us know that an image is upright when it is formed. As a rule, light typically moves rightward towards and through a lens, and it continues moving rightward after it passes through the lens. An object distance that is left of a lens is taken to be positive, and an image distance to the right of a lens is assigned a positive value. Substitution of two positive distance values into the equation above yields an inverted image. Somewhat similarly, the height of an upright object is assigned a positive value, whereas the height of an inverted image is assigned a negative value; thus, an inverted image would be assigned a ( – h_i ) value using the equation above. In this way, the algebra that relates height and distance within the system at hand yields physically consistent results.

As useful as the aforementioned relationships are, there will be times when the only physical parameters known are the object and image distances at hand and the radii of curvature ( R ) assigned to a given lens. The radius of curvature describes the radius value of two hypothetical spheres that overlap to create a thin-lens system: 

The first spherical surface encountered by a light ray traveling from left to right is that of the sphere situated to the right of the lens. This sphere’s radius ( r₁ ) is assigned a positive value. The surface from which a ray of light leaves is associated with the leftmost sphere, and it is assigned a negative value. When all is said and done, all interfaces that bulge toward the left have a positive radii value, and all interfaces that bulge to the right have negative values. In regard to the Thin-Lens Equation ( a.k.a. Lensmaker’s Formula ), the physical circumstance desired to exist outside of a lens is related to the index of refraction and radii of curvature that a thin lens must have to create the desired outcome sought after.

Q: A small lightbulb is situated on a central axis 120 cm leftward of a thin bi-convex lens. This lens has radii of curvature of 60 cm and 30 cm. If the lens has an index of refraction ( n ) value of 1.50, where will the corresponding image be located? Furthermore, is the image location different when the lens’ surface is flipped around? Unless otherwise indicated, assume that the lens is surrounded by air, which has an index of refraction value of 1.

Note: The index of refraction ( n ) is a ratio of the speed of light in a vacuum vs. its speed within some medium:

n = ( c / v )

Generally speaking, a high refractive value indicates that light will be bent to a greater extent than it would if the value of ( n ) were lessened. We solve the problem by plugging the values that have been given into the Lensmaker’s Formula:

( 1 / d_o ) + ( 1 / d_i ) = ( n – 1 )[ ( 1 / R₁ ) – ( 1 / R₂ ) ]

( 1 / 1.20 m ) + ( 1 / d_i ) = ( 1.50 – 1 )[ ( 1 / 0.60 m ) – ( 1 / – 0.30 m ) ]

( 1 / d_i ) = ( 0.50 )[ ( 1 / 0.60 m ) – ( – 2 / 0.60 m ) ] – ( 1 / 1.20 m )

( 1 / d_i ) = ( 0.50 )( 3 / 0.60 m ) – ( 1 / 1.20 m )

( 1 / d_i ) = ( 0.50 )( 1 / 0.20 m ) – ( 1 / 1.20 m )

( 1 / d_i ) = ( 2 / 1.20 m )

d_i = 0.60 m

The image distance is positive, so it lies to the right of the lens at hand. And surprisingly, if the lens values are reversed, the physical outcome is the same!!!

Try it out: 

( R₁ = 0.30 m ) and ( R₂ = – 0.60 m )