Question

Prove that R(radius of curvature)=2F(focal length).

Solve this question without using FP=FS as it is only possible for mirror of small curvature. (See attachment)

Answer 1

R = 2F is not accurate, it is valid only for small values of incident angle/aperture.

Drop a ⊥ from P on FS, name that O.   For small aperture

           FO = FS = focal length(f)

           CO = CS = Radius(R)

Let ∠QPC = θ = i = ∠CPF = θ = r

Using simple geometry(corresponding & alternate angles)

∠PCF = θ,        ∠PFS = 2θ,  

∴ tan∠PCF =  PO/CO

  tanθ  = PO/CO

  [tanθ≈θ, for small value of θ]

 ∴  θ = PO/CO    ...(1)

Similarly, in ΔPFO,

  2θ = PO/FO     ...(2)

Comparing (1) and (2):

⇒ 2(OP/CO) = PO/FO

⇒ 2FO = CO

⇒ 2F = R           [above, FO=f, CO=R]

     proved

Answer 2

Answer:

Required Answer :-

Let us assume that a ray of light is AB through the Let the principal axis incident of the mirror be at B. Now,

The normal surface point B are

CP

CB

Where

Length of CP = Length of CB = Radius of curvature

According to the question

AB will pàss through the Focal length. So, By using the law of reflection

$\bf \angle \: i = \angle \: r$

Now,

$\bf \angle BCP = \theta = i$

Now,

I = r

$\bf \angle BF = FC$

Now,

According to the question

The point B is close to P

FC = FP = PF

PC = PF + PC

Earlier we have discussed PC = PF

PC = 2PF

So,

R = 2F