The convex mirror shown in the drawing forms a virtual image of an arrow at x =

Question

The convex mirror shown in the drawing forms a virtual image of an arrow at x = x2 = 22.6 cm. The image of the tip of the arrow is located at y = y2 = 7.6 cm. The magnitude of the focal length of the convex mirror is 48 cm.

1)

What is x1, the x co-ordinate of the object arrow?.cm

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2)

What is y1, the y co-ordinate of the tip of the object arrow?cm

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3)

The object arrow is now moved such that image distance is halved, i.e., ximage,new = 11.3 cm. What is x1,new, the new x co-ordinate of the object arrow?cm

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4)

What is y2,new, the y co-ordinate of the image of the tip of the arrow when the x co-ordinate of the object arrow is equal to x1,new?

Explanation / Answer

f = -48 cm

v = -22.6 cm (image is beyond the mirror)

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

1/-48 = 1/u + 1/-22.6

u = 42.7 cm infront of the mirror

so u = x1 = -42.7 cm

2) m = -v/u = -(-22.6/42.7) = 0.529

m = hi/ho

ho = hi/m = 7.6/0.529 = 14.362 cm

y1 = +14.362 cm

3) v = -11.3 cm (beyond the mirror)

1/f = 1/u + 1/v

1/-48 = 1/u + 1/-11.3

u = 14.78 cm infront of the mirror

x1_new = -14.78 cm

4) m = -v/u = -(-11.3/14.78) = 0.7645

m = hi/ho

hi = ho*m = 14.362*0.7645 = 10.981 cm

y2_nwe = +10.981 cm

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