how can i prove mirror formula?
kvnmurty:
plane mirror or convex mirror or concave mirror ?
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Let us take the case of concave mirror. See diagram.
Let P be the pole (geometrical center) of the mirror. Let us draw the axis of the mirror. Let us draw the focal point F and the center C at distances f and 2f from the pole on the principal axis.
Let us take a point object at A which is at distance u from P. To find its image draw a ray AM on to the mirror at angle y to the axis. Draw the normal to the mirror at M which passes through the center C. Now draw the reflected ray making the same angle x with the normal. Let it intersect the axis at A'. Draw another ray from object A along the axis which gets reflected back along the same path. It intersects A'M at A'. Hence A' is the image of A.
We want to find a relation between u and v (image distance from P),
Z = exterior angle of Triangle AMA' and hence z = y + 2 x or, z - 2x = y
z-x is the exterior angle of triangle AMC.
Tan y = MP' / AP' Tan z = MP' / A'P' tan z-x = MP' / CP'
When y, x, z are all small angles. Tangent is nearly same as the angle. Hence
y = MP' / AP' z = MP' / A'P' z-x = MP' / CP'
So we get x = MP'/AP' - MP'/CP'
Since y = z - 2x , MP' / AP' = MP' / A' P' - 2 [ MP'/AP' - MP'/CP' ]
Cancel MP'.
1 /AP' = 1 / A' P' - 2 / A'P' + 2 /CP'
Substitute CP' = 2f and A'P' = v and u = AP'. as PP' is very small and is negligible for small aperture mirror and small angles x, y and z.
1 / u = - 1 / v + 2 / 2f
1/u + 1/v = 1/f -- equation 1
We introduce a cartesian sign convention. The distances in the direction of the light ray are taken as positive and those in the opposite direction are taken as negative.
So u is -ve, v is negative and f is also negative. Then equation 1 is
1/-u + 1/-v = 1/-f => 1/u + 1/v = 1/f
The same equation can be proved for convex mirror also.
Let P be the pole (geometrical center) of the mirror. Let us draw the axis of the mirror. Let us draw the focal point F and the center C at distances f and 2f from the pole on the principal axis.
Let us take a point object at A which is at distance u from P. To find its image draw a ray AM on to the mirror at angle y to the axis. Draw the normal to the mirror at M which passes through the center C. Now draw the reflected ray making the same angle x with the normal. Let it intersect the axis at A'. Draw another ray from object A along the axis which gets reflected back along the same path. It intersects A'M at A'. Hence A' is the image of A.
We want to find a relation between u and v (image distance from P),
Z = exterior angle of Triangle AMA' and hence z = y + 2 x or, z - 2x = y
z-x is the exterior angle of triangle AMC.
Tan y = MP' / AP' Tan z = MP' / A'P' tan z-x = MP' / CP'
When y, x, z are all small angles. Tangent is nearly same as the angle. Hence
y = MP' / AP' z = MP' / A'P' z-x = MP' / CP'
So we get x = MP'/AP' - MP'/CP'
Since y = z - 2x , MP' / AP' = MP' / A' P' - 2 [ MP'/AP' - MP'/CP' ]
Cancel MP'.
1 /AP' = 1 / A' P' - 2 / A'P' + 2 /CP'
Substitute CP' = 2f and A'P' = v and u = AP'. as PP' is very small and is negligible for small aperture mirror and small angles x, y and z.
1 / u = - 1 / v + 2 / 2f
1/u + 1/v = 1/f -- equation 1
We introduce a cartesian sign convention. The distances in the direction of the light ray are taken as positive and those in the opposite direction are taken as negative.
So u is -ve, v is negative and f is also negative. Then equation 1 is
1/-u + 1/-v = 1/-f => 1/u + 1/v = 1/f
The same equation can be proved for convex mirror also.
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Mirror
formula is the relationship between
object
distance (u), image distance (v) and focal length (f).
A ray of light starting from A and incident on the mirror along AD parallel to the principal axis of the mirror and gets reflected from the mirror and passes through F.
Another ray of light incident along AP is reflected along PA’ such that angle APB=i= angle BPA’= r.
The two reflected ray meet at A’ which is a real image.
Therefore A’B’ is real, inverted image of AB formed by reflection from the mirror.
In Triangles BAC and A’B’C’
AB/A’B’ = CB/CB’ ……………. (i)
Now Triangle ABP and A’B’P’ are similar,
AB/A’B’ = PB/PB’ …………………. (ii)
From (i) & (ii) …
CB/CB’= PB/PB’ ……………….. (iii)
Measuring all distances from P,
We have CB= PB-PC CB’= PC-PB’
Therefore from (iii),
(PB-PC)/(PC-PB’) = PB/PB’ …………… (iv)
Using new cartesian sign convections,
PB= -u (distance of object) PC= -R PB’ = -v (distance of image) We get from (iv), (-u+R)/(-R+v) = -u/-v
Or,
+uR – uv = uv – vR Or,
uR + vR = 2uv
Dividing both sides by uvR,
We get, 1/v + 1/u = 2/R As
R=2f
Therefore, 1/v + 1/u = 2/R = 2/2f = 1/f This is the required formula.
A ray of light starting from A and incident on the mirror along AD parallel to the principal axis of the mirror and gets reflected from the mirror and passes through F.
Another ray of light incident along AP is reflected along PA’ such that angle APB=i= angle BPA’= r.
The two reflected ray meet at A’ which is a real image.
Therefore A’B’ is real, inverted image of AB formed by reflection from the mirror.
In Triangles BAC and A’B’C’
AB/A’B’ = CB/CB’ ……………. (i)
Now Triangle ABP and A’B’P’ are similar,
AB/A’B’ = PB/PB’ …………………. (ii)
From (i) & (ii) …
CB/CB’= PB/PB’ ……………….. (iii)
Measuring all distances from P,
We have CB= PB-PC CB’= PC-PB’
Therefore from (iii),
(PB-PC)/(PC-PB’) = PB/PB’ …………… (iv)
Using new cartesian sign convections,
PB= -u (distance of object) PC= -R PB’ = -v (distance of image) We get from (iv), (-u+R)/(-R+v) = -u/-v
Or,
+uR – uv = uv – vR Or,
uR + vR = 2uv
Dividing both sides by uvR,
We get, 1/v + 1/u = 2/R As
R=2f
Therefore, 1/v + 1/u = 2/R = 2/2f = 1/f This is the required formula.
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