1. A ray coming from object
hits the concave mirror at a height at point A .
2. After reflection passes
through point I which is on the principal axis.
3. Here AC is the normal, and
here angle of incidence and angle of reflection
Are equal and are denoted by θ
4. Three right angle triangles
we can observe from the figure.
They are ΔACP', ΔAIP'.
5. Let the angle at vertices
O,C,I are α, β,γ.
6. We know, sum of the interior
angles is equal to exterior angles.
Δ AOC, β = α + θ
θ= β - α ------ (1)
Δ ACI,
γ = β + θ
γ = β + β - α
2β = α + γ ------(2)
When h becomes very small, P'
is taken as P
P'O = PO, P'C=PC, P'I= PI
In right angle triangle, Tan
value is the acute angle which is the ratio
Of opposite side toothed adjacent
side.
Tan α = P'A/P'O = h/PO
Tan β = P'A/P'C = h/PC
Ran γ = P'A/P'I= h/PI
When angle θ is very small
then Tanθ value become close to θ.
α = h/PO, β = h/PC, γ = h/ PI
Substitute in equation (2)
2h/PC = h/PO + h/PI
2/PC = 1/PO + 1/PI ------(3)
Sign Convention:
1. All the distance should be
measured from Pole.
2. The distance is measured
in the direction of incident ray is taken as positive, for opposite direction
is taken as negative.
3. Height is taken as positive
when measured upwards and taken as negative when measured downwards.
Here PC is
radius of curvature, taken as ‘-R’ (opposite to incident ray)
PO is object distance, taken as ‘-u’
PI is image distance, taken as ‘-v’
2/R = 1/u + 1/v
Since radius of curvature is
twice the focal length(R=2f)
2/2f = 1/u + 1/v
Mirror Equation: [1/f = 1/u + 1/v]
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