An object is p = 34.3 cm in front of a concave mirror. Its real image height is 6 times larger than the object height.
What is the location of the image?
Answer in units of cm.
What is the radius of curvature of the mirror? Answer in units of cm.
3qoweo
To find the location of the image formed by a concave mirror, we can use the mirror equation:
1/f = 1/di + 1/do
where f is the focal length of the mirror, di is the distance from the mirror to the image, and do is the distance from the mirror to the object. Rearranging the equation, we have:
di = 1/(1/f - 1/do)
Given that the object is located p = 34.3 cm in front of the mirror, we can substitute this value into the equation:
di = 1/(1/f - 1/p)
Since the problem does not provide the focal length of the mirror, we need to find it using the magnification equation:
magnification = -di/do
where magnification is the ratio of the image height to the object height. Given that the image height is 6 times larger than the object height, we have:
magnification = hi/ho = 6
Substituting -di for hi and -do for ho in the equation, we get:
magnification = -di/-do = di/do = 6
Now we have two equations:
di/do = 6
di = 1/(1/f - 1/p)
We can solve these equations simultaneously to find the values of di and f.
By substituting di = 6do into the second equation, we get:
6do = 1/(1/f - 1/p)
Simplifying the equation:
6do = p/(p - f)
Now, substituting the given values into the equation:
6do = 34.3 cm / (34.3 cm - f)
To find the value of f, we need another equation. The mirror equation can be expressed in terms of the radius of curvature (R) as:
1/f = 2/R
Solving this equation for f, we have:
f = R/2
Substituting this into the equation above:
6do = 34.3 cm / (34.3 cm - R/2)
We still have two variables (do and R), so we need more information to find the values of do and R.
Please provide additional information or values to proceed with the calculation.