A recent exhibit at the Franklin Institute was titled Galileo, the Medici and th

Question

A recent exhibit at the Franklin Institute was titled Galileo, the Medici and the Age of Astronomy. In the exhibit was an actual telescope used by Galileo. See Figure 1 for a schematic of a Galilean telescope. The objective lens is closest the object: the ocular lens is the lens the observer looks through. Note that the objective lens is convex while the ocular is concave. Denote the length of the telescope by 1 Suppose that F = +50.0 cm. An object with height h = 2.00 cm is placed on the optic axis of the telescope at p 1 = 75.0 cm from the objective lens. Ignoring the ocular lens, find: the image distance i 1 and the image height h 1. Is this image real or virtual? Upright or inverted? Suppose that f -10.0 cm(so that L = 40.0 cm). The image of the objective lens now serves as a virtual object for the ocular lens. For the ocular, find: the object distance p2, the image distance i 2 and the image height h 2. Is this image teal or virtual? Upright or inverted? (Hint: p 2

Explanation / Answer

The objective focal length F = 50 cm The height of the object is h1 = 2 cm The object distance is P1 = 75 cm A) The thin lens equation is 1/F = 1/P1+ 1/I1 I1 is the image distance fron the lens 1/50 = 1/75 + 1/I1 I1 = 150 cm the magnification m1 = -I1/P1 m1 = -150/75 = -2 The height of the image is h1' = 2h1 = 4 cm The image is real, inverted and erected B) The above image acts like the object (virtual) for the ocular lens The focal length of ocular is F2 = -10 cm Length of the tube is L = 40 cm The object distance for the ocular is P2 = 150 cm - 40 cm = 110 cm [here P2 = 40-150 = -110 cm can also be taken.(virtual image) Then the focla length of the ocular should be taken positive] The thin lens equation is 1/F2 = 1/P2 + 1/I2 1/-10 = 1/110 + 1/I2 I2 = -9.167 cm The magnification is m2 = -(-9.167)/110 = 0.083 The height of the image is h2' = m2h1' h2' = 0.332 cm The image is upright, minimized and virtual with respect to the image formed by the objective C) overall magnification is M = h2'/h1 = 0.332/2 = 0.166 I2 = -9.167 cm The magnification is m2 = -(-9.167)/110 = 0.083 The height of the image is h2' = m2h1' h2' = 0.332 cm The image is upright, minimized and virtual with respect to the image formed by the objective C) overall magnification is M = h2'/h1 = 0.332/2 = 0.166

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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