A convex mirror has a radiusof curvature of magnitude is | R | = 308 cm. An obje
ID: 1756787 • Letter: A
Question
A convex mirror has a radiusof curvature of magnitude is |R| = 308 cm. An object ofheight H1= 0.83 cm is placed on the opticalaxis a distance s1 =154 cm from themirror.
Before doing any calculations, draw all three principalrays on a printout of this problem or on a fresh piece of paper.This will give you an invaluable qualitative understanding of wherethe image is located, its orientation, etc.
(a) At what distance from the mirror is the image formed?Indicate with a positive answer a point in front of the mirror(left of mirror) and by a minus a point behind the mirror (right ofmirror).
s1' = cm
(b) Is the image real or virtual?
Answer as follows: Real Image = (1), Virtual Image =(2).
Image type =
(c) Calculate the magnitude of the image height.
H1' = cm
(d) Is the image upright or inverted?
Answer as follows: Upright Image = (11), Inverted Image =(22).
Image orientation =
The object is removed and instead convergingrays enter from the left. These rays would form a realupright image at a distanced = 77 cmbehind the convex mirror if the mirror wereabsent. (This situation is called a virtualobject.)
(e) With the mirror present, at what positions2'is an image formed?
s2' = cm
(f) Is the image in part (e) real or virtual?
Explanation / Answer
Given : Radius of curvature (R) = 308 cm Focal length (f) = - R /2 = - 308 / 2 = - 154 cm Object distance (u) = 154 cm We have : 1 / f = 1/ u + 1 / v 1/ v = - 1 / 154 - 1 /154 or Image distance (v) = - 77 cm (b) Image is virtual (c) We knowthat : Magnification (M ) = - v / u = hi / ho or 77 / 154 = hi / 0.83 hi = 0.415 cm (d) Image is Inverted. (e) In this situation : - 1 / u + 1/ v = 1/ f or 1 / u = - 1 / f + 1 / v = 1 / 154 + 1 / 77 u = 51.33 cm (f) Image is real Hope this helps u!Related Questions
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