In this example we will find the position and magnification of an image formed b

Question

In this example we will find the position and magnification of an image formed by a spherical mirror. A lamp is placed 10 cm in front of a concave spherical mirror that forms an image of the filament on a screen placed 3.0 m from the mirror. What is the radius of curvature of the mirror? What is the lateral magnification? If the lamp filament is 5.0 mm high, how tall is its image? SOLUTION SET UP (Figure 1) shows our diagram. SOLVE Both object distance and image distance are positive; we have s=10cm and s'=300cm. To find the radius of curvature, we use the following: 1s+1s110cm+1300cm==2R2R and R=19.4cm. The lateral magnification m is obtained from m=y/y=s/s, m=yy=150mm5mm=30 To find the height of the image, we simply multiply the height of the object by the magnification: y=30×5.0mm=150mm REFLECT The image is inverted (as we know because m=30 is negative) and is 30 times taller than the object. Notice that the filament is not located at the mirror’s focal point; the image is not formed by rays parallel to the optic axis. (The focal length of this mirror is f=R/2=9.7cm.) Part A - Practice Problem: A concave mirror has a radius of curvature R=25cm. An object of height 4 cm is placed 15 cm in front of the mirror. What is the image distance? Express your answers in centimeters to two significant figures. 75 cm SubmitMy AnswersGive Up Correct Part B - Practice Problem: What is the height of the image? Express your answer in centimeters to two significant figures.

Explanation / Answer

here s = 15 cm

focal length f = R/2 = 25 / 2 = 12.5 cm

mirror formula - 1/f = 1/s' + 1/s

so, 1/ 12.5 = 1/s' + 1/15

s' = 15 x 12.5 / ( 15 - 12.5) = 75 cm

Magnification m = - s'/s = -75 / 15 = -5

m = y' / y

object height y is given as 4cm

-5 = y' / 4

y' = - 20 cm -ve sign indicates an inverted image.

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