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 = 22.8 cm. The image of the tip of the arrow is located at y = y2 = 8.3 cm. The magnitude of the focal length of the convex mirror is 40 cm.

1)

2)

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

3)

The object arrow is now moved such that image distance is halved, i.e., ximage,new = 11.4 cm. What is x1,new, the new x co-ordinate of the object arrow?

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

Part 1)

For a convex mirror the focal length is negative

1/f = 1/p + 1/q

1/-40 = 1/p + 1/-22.8

p = 53.0 (This is x1)

Part 2)

M = -q/p = h'/h

8.3/h = -(-22.8)/53

h = 19.3 (This is y1)

Part 3)

1/-40 = 1/p + 1/-11.4

p = 15.9 cm (This is the new x1)

Part 4)

h'/19.3 = -(-11.4)/15.9

h' = 13.8 cm (The new y2)

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