Geometric Optics

Problem : Where is the image of an object placed 7 centimeters away from a 5 centimeter focal length convex lens? Concave mirror? Are the images real or virtual in each case?

Problem : Determine whether the images of the objects placed at the following positions will be upright or inverted. Concave mirror, s _o = 2f ; convex mirror, f < s _o < 2f ; concave lens, f > s _o .

Solution for Problem 2 >>

The orientation of the image is determined by the sign of the transverse magnification M _T = . For a concave mirror the focal length is positive. Thus 1/f = 1/2f + 1/s _i , and since 2f > f we have 1/f > 1/2f and s _i must be positive, so the magnification must be negative (the object and image distances are both positive). Hence the first image is inverted. For a convex mirror the focal length is negative, thus -1/f = 1/s _o +1/s _iâá’1/s _i = - (1/f + 1/s _o) , where f and s _o are now considered positive quantities. Thus s _i must be negative, the magnification positive and the image upright. In the third case, concave lenses have a negative focal length, thus -1/f = 1/s _o +1/s _i , and we have the same situation as before, so the image is upright.

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Problem : Determine whether the images of the objects placed at the following positions will be enlarged or diminished. Concave mirror, s _o = f ; convex mirror, f < s _o < 2f ; concave lens, s _o > 2f . Also determine whether the images are real or virtual.

Solution for Problem 3 >>

The magnification of the image is determined by the magnitude of the transverse magnification M _T = . For a concave mirror the focal length positive, thus: 1/f = 1/f + 1/s _iâá’1/s _i = 0 . This implies that the image forms at an infinite distance--in other words there is effectively no image. A convex mirror has a negative focal length, so -1/f = 1/s _o +1/s _iâá’ - (1/f + 1/s _o) = 1/s _i . Since s _o > f , 1/f > 1/s _o , thus the left hand side is positive. But this clearly implies that | 1/s _I| > | 1/s _o| and that | s _o| > | s _i| and thus the image must be diminished. s _i is negative so the image is virtual. In the last case, concave lenses have negative focal lengths so -1/f = 1/s _o +1/s _iâá’ - (1/f + 1/s _o) = 1/s _i . The term in parentheses is clearly greater than 1/s _i , so | s _i| < | s _o| so the image is diminished. Once again s _i is negative so the image is virtual.

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Problem : Use what you know about geometric optics to decide what type of lens is used in a magnifying glass and describe how it works.

Problem : A diagram of a Keplerian telescope is shown in the figure below. Note that the objective lens and eyepiece lens share a focal plane. The image formed by the objective must be in the region beyond the focal plane of the eyepiece lens. Compute the transverse magnification of such a telescope in terms of the focal lengths of the two lenses.

Figure %: A ray diagram of a Keplerian telescope.

Solution for Problem 5 >>

If we define x _i for a convex lens as the distance between the image and the focus on the near side of the lens, and x _o as the distance between the object and the nearest focus (there are foci on both sides of a convex lens), then | M _T| = = = (the second equality holds because triangles F _o OB and S ₁ S ₂ F _o are similar.

Figure %: Geometry of a convex lens.

Also, however, we can write | M _T| = = = since triangles AOF _e and F _e P ₁ P ₂ are similar. The magnification due to the objective lens is then M _o = x _i f _o , and the magnification due to the eyepiece lens is M _e = f _e x _o . But in this case it is clear from the diagram that x _i = x _o thus the total magnification is M = M _e M _o = .

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