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 = 19.9 cm. The image of the tip of the arrow is located at y = y2 = 6.6 cm. The magnitude of the focal length of the convex mirror is 39 cm.

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

2)What is y1, the y-coordinate of the tip of the object arrow?

3)The object arrow is now moved such that image distance is halved, i.e., ximage,new = 9.95 cm. What is x1,new, the new x-coordinate of the object arrow?

4)What is y2,new, the y-coordinate of the image of the tip of the arrow when the x-coordinate of the object arrow is equal to x1,new?

Explanation / Answer

In the first case,
Take f as the focal length, v as the image distance and u as the object distance.
f and v are negative for convex mirror.
Using the formula,
- 1/f = - 1/v + 1/u
1/u = 1/v - 1/f
= 1/19.9 - 1/39
1/u = 0.02461
u = 40.63 cm.

Magnification = - v/u
- v/u = hi/ho
hi and ho are image height and object height respectively.
- ( -19.9/40.63) = 6.6/hi
hi = 13.48 cm.
- - - - - - - - - - - - - - - - - - - - - - - -
In the second case,
v = - 9.95 cm
Using the formula 1/f = 1/u + 1/v,
1/u = 1/v - 1/f (By applying sign conventions)
= 1/9.95 - 1/39
u = 13.36 cm.

Magnification = - v/u
- v/u = hi/ho
hi and ho are image height and object height respectively.
- ( -9.95/13.36) = 6.6/hi
hi = 8.86 cm.

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