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a real image three times as large as the object is formed using a concave mirror of...
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Here, m = -3 (real image),
f = -20/2 = -10 cm (concave mirror).
Applying mirror equation and the magnification equation for concave mirror,
1/f = 1/d_o + 1/d_i
m = h_i/h_o = - d_i/d_o
Putting the values, we get
-1/10 = 1/d_o + 1/d_i ---- (i)
and -3 = - d_i/d_o ---- (ii)
or, d_i = 3 d_o
-1/10 = 1/d_o + 1/(3*d_o)
Or, (3+1)/(3d_o) = -1/10
Or, 4/(3d_o) = -1/10
Or, d_o = -40/3 = -13.33 cm (object on the left of the mirror)
d_i = 3*d_o = -40*3/3 = -40 cm (image on the left of the mirror).
Therefore the object was in the front of the image, 13.33 cm away from it, the image was also in the front side, 40 cm away from the concave mirror.
Posted by llltkl on June 5, 2013 at 4:28 PM (Answer #1)
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