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An object 9.00 mm tall is placed 12.0 cm to the left of the vertex of a convex mirror whose radius of curvature has a magnitude of 20.0 cm.(a) calculate the position, size, orientation (erect or inverted), and nature (real or virtual) of the image. (b) draw a principal-ray diagram showing the formation of the image (how would i do that.. i know you can't draw it here)..

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Answer:

Step-by-step explanation:

a )

for convex mirror, focal length

f = + 20 / 2 = 10 cm

object distance u = 12 cm ( negative )

Using mirror formula


(1)/(v) + (1)/(u) = (1)/(f)

Putting the values


(1)/(v) - (1)/(12) = (1)/(10)


(1)/(v) =(1)/(10) + (1)/(12)


v=(12* 10)/(12+10)

v = 5.45 cm

Size of image

=
(v)/(u) * size of object


=(5.45)/(12) * 9

= 4.08 mm .

It will be erect because only erect image is formed in convex mirror .

It will be virtual .

b )

Draw ray from tip of object parallel to principal axis and after reflection appears to be coming from focus . Second ray falls on the pole and reflects back making equal angle with principal axis . Draw the reflected ray backwards to cut the other ray . They cut each other where image is formed .

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