Why are two images possible with the fixed distance between mesh and screen? A f

Question

Why are two images possible with the fixed distance between mesh and screen? A full credit answer will include:

The two ray diagrams that you drew drew for procedures 1b and 1c, including numerical estimates for d0, di, and f in both cases.
A 2+ sentence explanation that references the two ray diagrams and explains the numerical connection that makes the two set-ups so similar.

Background:

Other materials: plastic ruler, meter stick, supports and lens mounts, double convex lens, double concave lens, wire mesh, cardboard screen, small plug-in lamp.
General concepts: the bending of light as it crosses from one medium to another (air to glass, or glass to air) is called refraction. Different colors of light are refracted to different degrees, a fact that we will ignore in this experiment. Light rays from a distant source are considered parallel; these rays are refracted toward a central axis by a converging lens and away from the axis by a diverging lens. In the case of parallel rays incident on a converging lens, the refracted rays form a real image at a focal point and we can easily measure the distance from the center of the lens to that point: that distance is what we call the focal length of the lens. In the case of a diverging lens, the parallel rays are interpreted as having originated from a virtual image one focal length behind the lens but we can’t easily locate that image and measure that distance. The focal length of a diverging lens is instead measured by examining the degree to which it “pries apart” the converging rays from a converging lens and causes the real image to form in a different position. This is in fact the principle behind the correction of nearsighted vision (myopia).
We will make our measurements using an arrangement called an optical bench: a meter stick with some attached support feet on which we can mount lens and screen supports. For accurate results, be sure that the lens and screen supports are installed perpendicular to the stick.

Explanation / Answer

focus image of an object at infinite by using, the lens. now the distance between lens and screen is equal to the rough focal lenght of the lens.

Place the screen at a distance of four times the focla lenght of the lens. Now move the lens from the mesh to screen. At a location of distnace d0 from the mesh, a enlarged image of the mesh forms on the screen. note the value of d1. agains move the lens towards the mesh, again a diminished clear image of mesh forms on the mesh. now note the distance d2 of the lens from the mesh.

The image distnace in the first case is,

v= 4f-d0

The image distance in the second case is also equal to v.

v'=d0

the object distnace in the first case is, d0 and the object distance in the second case is 4f-d0

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.