proove that angle of reflection is equal to the angle of incidence?
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As per law of reflection the angle of incidence is always equal to angle of reflection. This can be proved mathematically and also by a simple experiment. There are many mathematical proofs but the simplest proof is based on Fermat’s Principle.
Look at the above diagram. Let a right ray AO from a source A strike a plane mirror at O and reflected as ray OB. Let B be a point on ray OB and let ON be the normal to the mirror at O. The angle of incidence is ‘x’ and angle of reflection is ‘y’, as shown. Let ‘t’ be the time taken for the ray to travel the distance AO + OB and ‘c’ be the velocity of light, which is a constant.
The time ‘t’ can be worked out as,
t = AOc + OBc or,
t = √a2+p2c+√(l−a)2+q2c
As per Fermat's principle light takes the path of least time. Therefore, the derivative dtda must be 0.
That is, 0 = 1c[a√a2+p2+−(l−a)√(l−a)2+q2]
or, a√a2+p2 = (l−a)√(l−a)2+q2
But as per trigonometry, this is same as sin(x) = sin(y) or x = y.
Hence,
The Angle of Incidence = Angle of Reflection
The law of reflection can also be proved experimentally. The arrangement for the experiment is described below.
Take a white card board sheet and fix the same firmly on a board. Draw a thick line MM’ and mark the midpoint O. Draw a perpendicular ON at O. Place a plain mirror on MM’ with its center at O.Take three pins and fix them vertically on the sheet at positions C, B and A on one side of ON such that the pins lie on a straight line when viewed from A. Now move to the other side of the perpendicular ON and fix three more pins at F, E and D in such way that the images A’ B’ C’ of the pins at A, B, C and the pins at D, E, F are all in one straight line. This is the most important point of the entire experiment.
Now remove the mirror and the pins carefully marking the points where they have been fixed. At this stage we will not be able to locate the positions of A’, B’ and C’. Draw a perpendicular AP from A and extend that line. Join DO and extend that too. Let A’ be the point of intersection of these extended line. A’ is nothing but the position of the image that appeared when the mirror was there. Same way locate the positions of B’ and C’.
If you join AO and measure the angles AON and DON, you will find them to be equal. Thus, it proves the angle of incidence equals angle of reflection. Also you will notice that the following measures to be equal. AP = PA’, BQ = QB’ and CR = RC’, which proves the image distance equals the object distance.
Look at the above diagram. Let a right ray AO from a source A strike a plane mirror at O and reflected as ray OB. Let B be a point on ray OB and let ON be the normal to the mirror at O. The angle of incidence is ‘x’ and angle of reflection is ‘y’, as shown. Let ‘t’ be the time taken for the ray to travel the distance AO + OB and ‘c’ be the velocity of light, which is a constant.
The time ‘t’ can be worked out as,
t = AOc + OBc or,
t = √a2+p2c+√(l−a)2+q2c
As per Fermat's principle light takes the path of least time. Therefore, the derivative dtda must be 0.
That is, 0 = 1c[a√a2+p2+−(l−a)√(l−a)2+q2]
or, a√a2+p2 = (l−a)√(l−a)2+q2
But as per trigonometry, this is same as sin(x) = sin(y) or x = y.
Hence,
The Angle of Incidence = Angle of Reflection
The law of reflection can also be proved experimentally. The arrangement for the experiment is described below.
Take a white card board sheet and fix the same firmly on a board. Draw a thick line MM’ and mark the midpoint O. Draw a perpendicular ON at O. Place a plain mirror on MM’ with its center at O.Take three pins and fix them vertically on the sheet at positions C, B and A on one side of ON such that the pins lie on a straight line when viewed from A. Now move to the other side of the perpendicular ON and fix three more pins at F, E and D in such way that the images A’ B’ C’ of the pins at A, B, C and the pins at D, E, F are all in one straight line. This is the most important point of the entire experiment.
Now remove the mirror and the pins carefully marking the points where they have been fixed. At this stage we will not be able to locate the positions of A’, B’ and C’. Draw a perpendicular AP from A and extend that line. Join DO and extend that too. Let A’ be the point of intersection of these extended line. A’ is nothing but the position of the image that appeared when the mirror was there. Same way locate the positions of B’ and C’.
If you join AO and measure the angles AON and DON, you will find them to be equal. Thus, it proves the angle of incidence equals angle of reflection. Also you will notice that the following measures to be equal. AP = PA’, BQ = QB’ and CR = RC’, which proves the image distance equals the object distance.
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Proof is given in the below image
➫here for x + y is always equal to 90°
As this is fundamental nature of light as when mirror is kept in such a position as stated
➫➫Whenever a ray of light i.e. the incident will hit a point at plane Generally kept in a(position as stated) mirror i.e. the normal then the ray of light will get reflected back in the same way i.e. the reflected ray. both rays form an angle on the normal i.e. the angle of incidence and angle of reflection. now as the angles formed, the imaginary line, i.e the normal, is a right angle i.e 90 degrees and both the angles of incidence and reflection respectively get divided into two 45 degrees angles respectively.
☄☄I hope it helps
➫here for x + y is always equal to 90°
As this is fundamental nature of light as when mirror is kept in such a position as stated
➫➫Whenever a ray of light i.e. the incident will hit a point at plane Generally kept in a(position as stated) mirror i.e. the normal then the ray of light will get reflected back in the same way i.e. the reflected ray. both rays form an angle on the normal i.e. the angle of incidence and angle of reflection. now as the angles formed, the imaginary line, i.e the normal, is a right angle i.e 90 degrees and both the angles of incidence and reflection respectively get divided into two 45 degrees angles respectively.
☄☄I hope it helps
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