The magnitude of the radius of curvature of the concave mirror in the figure is|
ID: 1756786 • Letter: T
Question
The magnitude of the radius of curvature of theconcave mirror in the figure is|R| = 41 cm. An object of height H1 =0.2 cm is placed upright on the optical axis at a distance ofs1 = 32.8 cm to the left of the mirror.
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. Make your ray drawingapproximately to scale and use a straight edge when drawing rays.It is not important that the curved face of the mirror be drawn toscale; in fact this usually complicates understanding rather thanhelping. Doing geometrical optics problems without drawing raydiagrams first is akin to going alone into the wilderness without amap (or GPS).
(a) At what distance from the mirror is the image formed?
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 =
(e) The object is now moved to a new locations2 and the resulting image is located at adistance behind the mirror equal in magnitude tothe distance found in part (a). Where has the object been placed toachieve this situation? Give a numerical value with a positive signdenoting in front of the mirror and a negative sign denotingbehind.
s2 = cm
(f) Calculate the magnitude of the heightH2' of the image of part (e).
Explanation / Answer
Given : Radius of curvature (R) = 41 cm Focal length (f) = 41 /2 = 20.5 cm Objectdistance (u) = 32.8 cm (a) We know that : 1 / f = 1/ u + 1/ v or 1 / v = 1/ f - 1 / u = 1 / 20.5 - 1 / 32.8 Image distance (v) = 54.7 cm (b) Image is real (c) We know that: Magnification = - v / u = - hi / ho or 54.7 / 20.5 = hi / 0.2 Imagedistance hi = 0.535 cm (d) Image isUpright. (e) As itis placed behind the mirror -1 / u + 1/ v = 1 / f - 1/ u + 1 / 54.7 = 1 / 20.5 or 1 /u = - 1 /20.5 + 1 / 54.7 Object distance is : u = - 32.79 cm (f) Magnification (M) = - v / u = - hi /ho or - 54.7 / 32.79 = hi / 0.2 cm hi = -0.334 cm Hope this helps u!Related Questions
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