A student notices the image produced from a large concave mirror of a distant sk

Question

A student notices the image produced from a large concave mirror of a distant skyscraper (h = 122m) is inverted and the magnitude of the height is |hi|=0.99m tall. The image is located d= 120 cm in front of the mirror. A) Write an expression for the distance between the mirror and the skyscraper, ds . B) Numerically, what is the distance in meters? C) Numerically, what is the focal point of the mirror,f, in cm?

Note: I have anser for part (B): 147.87m, however looking anser for part A & C

My answer for part A: ds=(d*h)/hi which incorrect,( Must use this character for answer: alpha / bita / theta / a / b / d / g / h / hi / j / k / m / P / S / t ) only.

My answer for part c: 84 cm also incorrect

Explanation / Answer

The mirror equation is
1/ f = 1/ i + 1/ p
Where i is the image distance and p is the object distance
The magnitude of the magnification
m = - i / p = h' / h
p = - i x (h / h')
ds  = - d x (h / h')
Where h' is the height of the image and h is the height of the object
b) Using the above equation
p = - i x (h / h')
p = - (-120 cm) x (122 m / 0.99 m)
p = 14814.81 cm
p = 148..15 m
c) Using the mirror equation
1 / f = 1 / i + 1/ p
1 / f = 1 /(-120 c m) + 1/ 148.15 m
1 / f = 1 / 14814.81 - 1 / 120 cm
1 / f = 6.75 x 10-5  - 833.33 x 10-5
1 / f = - 826.58 x 10-5 cm-1
f = - 0.001209804 x 105 cm
f = - 120.98 cm

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