The simplest gravitational lens is the lens around a distant star or planet. This lens is a point gravitational lens, because it behaves as though all of the mass of the gravitating object were concentrated at the very center of the lens. The structure of a point gravitational lens depends only on the mass of the central object, and the only role the radius of the central source plays is to set the minimum distance a lens must be from us to produce an image of a more distant object.

Light experiences the largest deflection as it passes through the gravitational lens of a star when the light just grazes the photosphere of the star. This defines a minimum focal length for the gravitational lens, or a minimum distance that an observer must be from a star to see it act as a lens. This distance is given by

where *D* is the distance from the observer to the star generating the lens,
*R* is the radius of the star, *c* is the speed of light, *G*
is the gravitational constant, and *M* is the mass of the star. The second
form of this equation gives the distance in Astronomical Units for the mass and radius
in units of the solar values.

For a main-sequence star, the radius of the star is very roughly proportional to the mass,
so the minimum distance one must be from a main sequence star to see its lens is about the
same for all stars. Given that the nearest star is over a
parsec
away, and a parsec is 2 × 10^{5} AU, potentially the gravitational lens
of every star in our Galaxy is visible to us.

For a solid planet, where the mass is proportional to *R ^{3}* and the
density varies relatively little from planet to planet, the minimum distance required
to see its gravitational lens is proportional to

The area on the sky covered by a gravitational lens is roughly the area encircled by
its Einstein ring. The larger this ring is on the sky, the more likely another more
distant object will be seen through the lens. But most of the stars in our own galaxy
are very far from us, and the Einstein ring of each of these stars covers a very small
area on the sky. A star at the edge of the plane of our galaxy is about 50 pc away from
us; if this star is one solar mass, its Einstein ring would cover only
4 × 10^{-15} of the sky. A more typical star in our galaxy is around 10 kpc
away from the Sun; if this star is of one solar mass, it would cover 10^{-19}
of the sky. Placing the 10^{12} solar masses worth of stars at 10 kpc, which
would constitute the total mass of our galaxy, we would find that altogether they cover
only about 10^{-7} of the sky. The chances that any one object outside of our
galaxy is behind the lens of a star within our galaxy is therefore only about one in ten
million.

An important point about this last calculation is that this probability depend only on
the distance to the lenses, and not the mass of each individual star. The radius of an
Einstein ring is proportional to *M ^{1/2}*, so the area is proportional to

A point gravitational lens produces only two types of image: either a single Einstein ring, or a pair of images. An Einstein ring is created whenever the image source lies on a direct line with the lens and us. If the source is away from this line, the lens creates two separate and distorted images of the source, one on each side of the lens. This second case is much more likely than the first if the size on the sky of the image source is much smaller than the radius of the Einstein ring.

The transition from ring to double image is easy to imagine. Let us assume that the source is a star in another galaxy and that the lens source is a star in our own Galaxy. We start with the image source in a direct line behind the lens. The ring that is produced is uniform all the way around the lens. As we move our extragalactic star to the left, we see the ring appear to shift to the left. The ring remains intact until the right edge of the extragalactic star is on the lens-observer line; when the extragalactic star's edge drops off of this line, the ring splits into two crescents, one facing left, the other facing right, with the points of one nearly touching the points of the other. As we move the star further to the left, the points of the crescent pull back and become rounded, until the crescents transform into two curved ovals, one on the left of the lens, the other on the right of the lens.

When the line to the extragalactic star falls inside the region defined by the Einstein ring, the lens creates images that are larger on the sky, and therefore much more luminous, than the unmagnified source. As our extragalactic star continues to move left, the images created by the lens continue to move left and shrink in size. Once the extragalactic star is outside of the area enclosed by the Einstein ring , the image on the right side of the lens becomes dimmer than the unmagnified source, and the image to the left of the lens becomes as bright as the unmagnified source. Eventually the image on the right of the lens becomes too dim to see, and the image on the left is the unmagnified image of the extragalactic star.

A star at 1 pc produces an Einstein ring of radius 0.29." A star of one solar mass at a more typical distance of 100 pc produce a ring of radius 0.03." The Hubble telescope can point to a position with a 0.01" resolution, so an Einstein ring is at the limit of detectability with this telescope. Realistically, we will never see the image created by a stellar gravitational lens.

The tell-tail signature of a point gravitational lens is its the brightening of more distant
stars. The maximum magnification is simply the area of the Einstein ring on the sky divided
by the area of the unmagnified image, which for a distant star is an increase in luminosity
of *2 α/φ*, where *α* is the radius of the Einstein ring on
the sky and *φ* is the radius of the unmagnified image on the sky. Consider for
example the magnification of a one solar mass star at 100 pc by the gravitational lens produced
by a one solar mass star at 10 pc. The star at 10 pc produces an Einstein ring with radius
α = 0.029." The radius on the sky of the star at 100 pc if the lens is absent is
φ = 4.7 × 10^{-5}." The gravitational lens can therefore increase the
luminosity of the 100 pc star by a factor of 1200.

The more likely outcome of a lens magnifying a more distant star is that the more distant
star falls near the circle defined by the Einstein ring. The probability of a star's image
having the maximum possible magnification to the probability that it is magnified at all
is roughly the area of the unmagnified star on the sky to the area enclosed by the Einstein
ring. For our last example, this probability is only 3 × 10^{-6}. The most
likely outcome of magnification is that the luminosity of the distant star increases by
less than a factor of two.