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 = 16.2 cm. The image of the tip of the arrow is located at y = y2 = 6.5 cm. The magnitude of the focal length of the convex mirror is 33 cm.

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

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 = 8.1 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

1) 1/p + 1/q = 1/f

1/p + 1/-16.2 = 1/-33

x1 = p = 31.82 cm

2)

M = y2/y1 = -q/p

y1 = 6.5*31.82/16.2= 12.77 cm

3)

1/p + 1/-8.1 = 1/-33
p=10.74 cm

4)
y2 = -q y1/p = 8.1*12.77/10.74= 9.63 cm

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