Confused about the formula
(1/-f)= (1/do)+(1/-di)
For locating an image for a convex mirror.
A rose is placed 15cm in front of a convex mirror of focal length 10cm. Determine the location of the image using the mirror equation.
(1/-10)=(1/15)+(1/-di)
-0.1=0.067 + (1/-di)
-0.166=(1/-di)
-di=(1/-0.167)
-di=-9.5941
di=9.5941
Am I on the right track there? I feel like I must be wrong bc that places the image before the focal point...
Thanks a bunch :)
draw a ray diagram to confirm your work.
I am & I can't get it to match up... They haven't given an exact height, they just say "at least 4 squares high" but between 4 squares & my cap I can't get it to that point.
You are on the right track in using the mirror equation to solve for the location of the image. However, there is a sign error in your calculation.
Let's go through the steps again:
The mirror equation is given by:
(1/-f) = (1/do) + (1/di)
Given that the focal length (f) is 10 cm, and the object distance (do) is 15 cm, we can substitute these values into the equation:
(1/-10) = (1/15) + (1/di)
Now, let's solve for the image distance (di):
(1/-10) = (1/15) + (1/di)
To simplify the equation, we need to find a common denominator. The common denominator in this case is 15 * di:
(di/-10) = (di/15) + (15/-10)
Multiplying both sides of the equation by the common denominator, we get:
(di * di) / -10 = (15 * di) + (-15 * (15))
Simplifying further:
(di * di) / -10 = 15di - 225
Multiply both sides of the equation by -10 to get rid of the fraction:
di * di = -150di + 2250
Rearranging the equation:
di * di + 150di - 2250 = 0
Now, we can solve this quadratic equation for di. You can use the quadratic formula or factoring to find the values of di. In this case, factoring is easier:
(di + 75)(di - 30) = 0
From this equation, we get two possible solutions for di: di = -75 or di = 30.
Since the image distance can't be negative for a convex mirror, the correct solution is di = 30 cm.
Therefore, the location of the image is 30 cm behind the convex mirror.
Note: The mirror equation assumes a positive value for the image distance (di) when the image is virtual or formed on the same side as the object. For a convex mirror, the image formed is always virtual and located on the same side as the object.