10pm.
I asked
"Show mathematically that to produce an image that is the same size as the object in a concave mirror, the object needs to be at twice the focal length from the mirror."
and bobpursley answered:
"1/o + 1/o=1/f
1/2f + 1/i=1/f
1/i=1/2f
i=2f
M=di/do= 2f/2f= 1
check my thinking."
could someone please explain, I'm not really following.
the mirror equation
1/do+ 1/di= 1/f I assume you are familiar with that. I see I had a typo in the first line.
multiply by do
do/do+do/di= do/f
1+ M=do/f
but M (do/di=ho/hi= 1 given)
1+1=do/f which means that do is twice f.
Sure! Let's go step by step to understand the answer provided by bobpursley.
Bobpursley started by using the mirror equation, which relates the object distance (o), the image distance (i), and the focal length (f) of a mirror. The mirror equation is written as:
1/o + 1/i = 1/f
The goal is to show mathematically that when the object distance is twice the focal length, the image produced by a concave mirror will be the same size as the object.
So, let's substitute the given object distance and find the image distance:
1/2f + 1/i = 1/f
To find i, we can combine the terms with a common denominator:
[1/(2f) + 1/i] = (1* i + 2f)/(2f * i) = (i + 2f)/(2f * i)
The equation now becomes:
(i + 2f)/(2f * i) = 1/f
Cross-multiplying gives:
(i + 2f) = 2f
To solve for i, we'll isolate the variable on one side:
i = 2f - 2f
Simplifying:
i = 0
This means that the image distance is zero. When the image distance is zero, the image is formed at infinity, and such an image is said to be "at infinity." Consequently, the image is the same size as the object.
Finally, bobpursley checked their thinking by calculating the magnification ratio (M). The magnification (M) is the ratio of the image distance to the object distance:
M = i/o = 0/(2f) = 0
Since M = 0, it confirms that the image size equals the object size.
In summary, when the object distance is twice the focal length, the image formed by a concave mirror is at infinity and therefore the same size as the object.