prove the plane mirror (360/n-1)
Answers
Answer:
This image shows you the top view of the mirrors and their images..
Let say actual mirrors OA and OB are kept at some angle ‘θ’.
So OB’ is the image of the mirror OB in the mirror OA.
Similarly OA’ is the mirror image of OA through mirror OB.
OB’’ is mirror image of mirror OB through the imaged-mirror OA’ and OA’’ is the mirror image of OA through imaged-mirror OB’. And it continues to other end (through out the 360 degrees).
The object here is represented by the big dot.
There will be images through OA and OB mirrors and again mirror images of these images through OA’ and OB’.
Again these mirror images have their images in the mirrors OA’’ and OB’’ and so on till that point where two mirror images gets coincide or their rays gets coincide.
In the above image, you can see there are two images at the top. If you try to find the possible location of the images of those two dots, then they may get coincided with each other or they may not form at all.
The images only coincide only when the angle (which is not kept blank) at the top is equal to θ.
The images do not form when the angle (which is not kept blank) at the top is less than θ.
So in general for any θ< 360 deg.
n(θ) + α = 360, where α is the angle at the top (which is left blank in the image).
If α = θ, it means that 360 is multiple of θ.
If α = θ, then (n+1)θ = 360.
n + 1 = (360/θ), where n+1 is equal to number of segments the whole 360 degrees is divided by the images of the mirrors.
In each segment one dot appears and out of them one of them is real. So total n+1 dots are formed and actual number of images is (n+1) -1 = n
so n = (360/θ)-1.
If α < θ, then kθ < 360 < (k+1)θ
k < 360/θ < k + 1.
Actually number of segments whole 360 degrees divided into, is equal to 360/θ which is not an integer. since α not equal to θ.
The number of segments with segment angle θ formed is equal to n. So number of images formed is equal to k-1.
Let say number of images formed is n= k-1, where k = [360/θ]
So number of images formed = [360/θ]-1 or [360/θ - 1]
Let say actual mirrors OA and OB are kept at some angle ‘θ’.
So OB’ is the image of the mirror OB in the mirror OA.
Similarly OA’ is the mirror image of OA through mirror OB.
OB’’ is mirror image of mirror OB through the imaged-mirror OA’ and OA’’ is the mirror image of OA through imaged-mirror OB’. And it continues to other end (through out the 360 degrees).
The object here is represented by the big dot.
There will be images through OA and OB mirrors and again mirror images of these images through OA’ and OB’.
Again these mirror images have their images in the mirrors OA’’ and OB’’ and so on till that point where two mirror images gets coincide or their rays gets coincide.
In the above image, you can see there are two images at the top. If you try to find the possible location of the images of those two dots, then they may get coincided with each other or they may not form at all.
The images only coincide only when the angle (which is not kept blank) at the top is equal to θ.
The images do not form when the angle (which is not kept blank) at the top is less than θ.
So in general for any θ< 360 deg.
n(θ) + α = 360, where α is the angle at the top (which is left blank in the image).
If α = θ, it means that 360 is multiple of θ.
If α = θ, then (n+1)θ = 360.
n + 1 = (360/θ), where n+1 is equal to number of segments the whole 360 degrees is divided by the images of the mirrors.
In each segment one dot appears and out of them one of them is real. So total n+1 dots are formed and actual number of images is (n+1) -1 = n
so n = (360/θ)-1.
If α < θ, then kθ < 360 < (k+1)θ
k < 360/θ < k + 1.
Actually number of segments whole 360 degrees divided into, is equal to 360/θ which is not an integer. since α not equal to θ.
The number of segments with segment angle θ formed is equal to n. So number of images formed is equal to k-1.
Let say number of images formed is n= k-1, where k = [360/θ]
So number of images formed = [360/θ]-1 or [360/θ - 1]