Question

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Derivation of Mirror Formula for a concave mirror when it comes to form a real image...
please explain it in a simple way...

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Answer 1

The diagram is given in the attachment .

The triangles AOB and A₁OB₁ are similar .

Hence :

BA/B₁A₁ = OB/O₁B₁

Similarly triangles DOF₂ and A₁ B₁ F₂ are similar .

So :

OD/B₁A₁ = OF₂/F₂B₁

Since BA = OD ,

= > OD/B₁A₁ = OF₂/F₂B₁

From 1 and 2 we get :

OB/OB₁ = OF₂/F₂B₁

Now F₂B₁  = OF₂ - OB₁

= > F₂B₁ = f - v

OF₂ = f

OB = u

OB₁ = v

When the image is real ,

u = -ve , v = +ve .

So we get :

(-u)/(v) = -f /( f - v )

= > u/v = f/(f - v )

= > uf - uv = fv

divide both sides by uvf :

= > 1/v - 1/f = 1/u

= > 1/f = 1/v - 1/u

Hence proved !

Answer 2

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Gdmrng☆▪☆

Mirror Formula: M1M2 is a concave mirror having pole P, focus F and centre of curvature C.

An object AB is placed in front of mirror with point B on the principal axis. The image formed by mirror is A' B' . The perpendicular dropped from point of incidence D on principal axis is DN.

In ΔABC and ΔA' B'C

∠ABC = ∠A' B'C (each equal to 90°)

∠ACB = ∠A'CB' (opposite angles)

Both triangles are similar.

Now in ΔDNF and A' B'F

∠DNF = ∠A'B'F (each equal to 90°)

∠DFN = ∠A'FB' (opposite angles)

∴ Both triangles are similar

If aperture of mirror is very small, the point N will be very near to P, so FN = FP

By sign convention

Distance of object from mirror PB = -u

Distance of image from mirror PB' = - v

Focal length of mirror PF = - f

Radius of curvature of mirror PC = - R = - 2f

Substituting these values in (4), we get

We get

1/f=1/u-1/v

Hope it helps u☆▪☆