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prove of mirror formula
Asked by fawadafzalkhan.on | 26 Oct, 2019, 03:58: PM
answered-by-expert Expert Answer
For getting image due to the reflection of spherical mirrors, we use the following guidelines in the ray diagram
 
(i) The ray from the point which is parallel to the principal axis. The reflected ray goes through the focus of the mirror.
(ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a
            convex mirror. The reflected ray simply retraces the path.
(iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through
       (or directed towards) the focus of a convex mirror. The reflected ray is parallel to the principal axis.
(iv) The ray incident at any angle at the pole. The reflected ray follows laws of reflection.
 
Figure shows the ray diagram considering three rays. It shows the image A′B′ (in this case, real) of an object AB formed
by a concave mirror.
 
We now derive the mirror equation or the relation between the object distance (u), image distance (v) and the focal length ( f ).
 
From Figure, the two right-angled triangles A′B′F and MPF are similar. (For paraxial rays, MP can be considered to be a straight line
perpendicular to CP.) Therefore,

begin mathsize 14px style fraction numerator B apostrophe A apostrophe over denominator P M end fraction space equals space fraction numerator B apostrophe F over denominator F P end fraction space space o r space space fraction numerator B apostrophe A apostrophe over denominator B A end fraction equals space fraction numerator B apostrophe F over denominator F P space end fraction space left parenthesis space b e c a u s e space P M space equals space A B right parenthesis end style  ...........................(1)
Since begin mathsize 14px style angle end style APB = begin mathsize 14px style angle end style A′PB′, the right angled triangles A′B′P and ABP are also similar. Therefore,
begin mathsize 14px style fraction numerator B apostrophe A apostrophe over denominator B A end fraction space equals space fraction numerator B apostrophe P over denominator B P end fraction end style.............................(2)
Comparing Eqs. (1) and (2), we get

begin mathsize 14px style fraction numerator B apostrophe F over denominator F B end fraction space equals space fraction numerator B apostrophe P space minus space F P over denominator F P end fraction space equals space fraction numerator B apostrophe P over denominator B P end fraction end style  .............................(3)
Equation (3 ) is a relation involving magnitude of distances. We now apply the sign convention.
We note that light travels from the object to the mirror MPN. Hence this is taken as the positive direction.
To reach the object AB, image A′B′ as well as the focus F from the pole P, we have to travel opposite to the
direction of incident light. Hence, all the three will have negative signs.
 
Thus,  B′ P = –v, FP = –f, BP = –u

Using these in Eq. (3), we get
 
begin mathsize 14px style fraction numerator negative v plus f over denominator negative f end fraction space equals space fraction numerator negative v over denominator negative u end fraction space space space o r space space space fraction numerator v minus f over denominator f end fraction space equals space v over u space space o r space space 1 over v space plus space 1 over u space equals space 1 over f end style
above relation is known as mirror equation
Answered by Thiyagarajan K | 26 Oct, 2019, 07:50: PM

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