A virtual image -7 cm tall and 17 cm away from a mirror is created by an object 7 cm away from a mirror. How tall is the object? What kind of mirror created the image? What is its focal length?
To find the height of the object, we can use the magnification equation:
magnification (m) = Height of image (hi) / Height of object (ho)
Given that the height of the virtual image (hi) is -7 cm (negative sign indicates it's an inverted image) and the distance of the object (ho) from the mirror is 7 cm, we can substitute these values into the magnification equation:
m = -7 cm / ho
Since m = hi / ho, we can rearrange the equation to solve for ho:
ho = hi / m
ho = -7 cm / m -------- (1)
Now, to find the magnification (m), we can use the mirror formula:
1/f = 1/do + 1/di,
where f represents the focal length of the mirror, do represents the object distance, and di represents the image distance.
Given that the image distance (di) is 17 cm and the object distance (do) is 7 cm, we can rearrange the mirror formula to solve for m:
1/f = 1/do + 1/di -------- (2)
Substituting the values into equation (2):
1/f = 1/7 cm + 1/17 cm
To simplify the equation, we can multiply each term by the denominator of the other term:
1/f = (17/17*7) + (7/7*17)
1/f = (17 + 7) / (17*7)
1/f = 24 / 119
Now, we can find the value of f by taking the reciprocal of both sides:
f = 119 / 24 cm
Using equation (1), we can substitute the value of m and solve for ho:
ho = -7 cm / (119 / 24 cm)
ho = -7 cm * (24 cm / 119)
ho ≈ -1.41 cm
Since the height cannot be negative, the absolute value of ho gives us the height of the object:
|ho| ≈ 1.41 cm
Therefore, the height of the object is approximately 1.41 cm.
Based on the given information and calculations, the image is created by a concave mirror since the height of the image is negative (-7 cm). The focal length of the mirror is approximately 119/24 cm.