A concave spherical mirror has a radius of curvature of magnitude 19.8 cm. (a) F

Question

A concave spherical mirror has a radius of curvature of magnitude 19.8 cm.     (a) Find the location of the image for the following object distances. (If there is no image formed enter "NONE".)
object distance (cm) image distance (cm) location 40.6 1 2 19.8 3 4 9.9 5 6
    (b) For each case, state whether the image is real or virtual.
object distance (cm) real/virtual 40.6 7 19.8 8 9.9 9
    (c) For each case, state whether the image is upright or inverted.
object distance (cm) real/virtual 40.6 10 19.8 11 9.9 12
     (d) Find the magnification in each case. (If there is no image formed enter "NONE".)
object distance (cm) magnification 40.6 13 19.8 14 9.9 15 A concave spherical mirror has a radius of curvature of magnitude 19.8 cm.     (a) Find the location of the image for the following object distances. (If there is no image formed enter "NONE".)
object distance (cm) image distance (cm) location 40.6 1 2 19.8 3 4 9.9 5 6
    (b) For each case, state whether the image is real or virtual.
object distance (cm) real/virtual 40.6 7 19.8 8 9.9 9
    (c) For each case, state whether the image is upright or inverted.
object distance (cm) real/virtual 40.6 10 19.8 11 9.9 12
     (d) Find the magnification in each case. (If there is no image formed enter "NONE".)
object distance (cm) magnification 40.6 13 19.8 14 9.9 15     (a) Find the location of the image for the following object distances. (If there is no image formed enter "NONE".)
object distance (cm) image distance (cm) location 40.6 1 2 19.8 3 4 9.9 5 6
    (b) For each case, state whether the image is real or virtual.
object distance (cm) real/virtual 40.6 7 19.8 8 9.9 9
    (c) For each case, state whether the image is upright or inverted.
object distance (cm) real/virtual 40.6 10 19.8 11 9.9 12
     (d) Find the magnification in each case. (If there is no image formed enter "NONE".)
object distance (cm) magnification 40.6 13 19.8 14 9.9 15 object distance (cm) image distance (cm) location 40.6 1 2 19.8 3 4 9.9 5 6

Explanation / Answer

The radius of curvature of the mirror R = 19.8cm

then the focal length f = R/2 = 9.9 cm

(a) From mirror formula

       1/f = 1/di + /1do

then the image distance

        di = fdo / do - f

            = (9.9)(40.6) / 40.6 - 9.8

             = 13.09cm

since the image distance is positive so the image is real

when the object distance 19.8cm

       di = (9.9)(19.8) / 19.8 - 9.9

           = 19.8 cm

the image real

whenthe object distance d0 = 9.9

      di = (9.9)(19.8) / 9.9-9.9 = (infinite)

   the image is real

Magnification m1 = - 13.09/40.6 = - 0.322

                       m2 = - 19.8 / 19.8 = -1

and                m3 = - infinite/9.9 = - inifinte

since the magnification is negative so all the images are inverted

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