capa14 - Solution Derivations for Capa #14 1) An image of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution Derivations for Capa #14 1) An image of the moon is focused onto a screen using a converging lens of focal length ( f = 34 . 8 cm ). The diameter of the moon is 3 . 48 × 10 6 m , and its mean distance from the earth is 3 . 85 × 10 8 m . What is the diameter of the moon’s image? f = Given h = Given (diameter of moon, height of object) l = Given (distance to object) The magnification equation can be used here: h 0 h = - l 0 l We are looking for h 0 , the height (diameter) of the image on the screen. Since the moon is so far away, we can treat the light rays as being parallel. Thus, the image distance l 0 is the focal length of the lens. h 0 = - hf l 2) A 0 . 54 cm high object is placed 8 . 5 cm in front of a diverging lens whose focal length is - 7 . 5 cm . What is the height of the image? h = Given l = Given f = Given We can’t use the magnification equation immediately because we don’t know the height of the image or the distance to it. But, the lens equation also relates these quantities. Thus, we can start with it and solve for the first unknown, the image distance. 1 l + 1 l 0 = 1 f 1 l 0 = 1 f - 1 l l 0 = 1 1 f - 1 l Now, we can use the magnification equation. h 0 h = - l 0 l h 0 = - hl 0 l 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
= - h l 1 1 f - 1 l = - h l ± 1 f - 1 l ² 3) A magnifying glass uses a converging lens with a focal length of 15 . 5 cm . It produces a virtual and upright image that is 2.7 times larger than the object. How far is the object from the lens? f = Given M = Given l = ? The magnification equation in another form is M = - l 0 l Thus, l 0 = - lM Now, using the lens equation, the object distance can be solved for 1 f = 1 l + 1 l 0 = 1 l - 1 lM = M lM - 1 lM 1 f = M - 1 lM lM = ( M - 1) f l = ( M - 1) f M 4) What is the image distance? (Think carefully about whether the answer is positive or negative.) The image distance formula was a necessary step for the final formula in #3. It is l 0 = - lM Since l was calculated in #3, simple plug this in. Note the signs, the algebra will take care of it. 5) In the 7 diagrams below, the solid arrow represents the object and the dashed arrow
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/03/2008 for the course PHYS 1120 taught by Professor Rogers during the Fall '08 term at Colorado.

Page1 / 6

capa14 - Solution Derivations for Capa #14 1) An image of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online