Deduce the relation , u2/v - u1/u= u2-u1/r when reflection takes place from rarer to denser medium(u2>u1) in case of spherical surface
Answers
Consider one medium with refractive index u1 and second medium with refractive index u2.
There is an object O and a ray of light from the object O is incident on the spherical mirror. Since it is moving from a rarer medium to a denser medium, the ray move towards the normal. An image is formed and radius of curvature of a spherical surface is 'r' with the center C of the spherical surface.
Since , we need to prove , u2/v -u1/u =(u2-u1)/r
where u1 =Refractive index of air
u2= Refractive index of glass
u= object distance
v=image distance
r= radius of curvature
Now,
By snell's law ,
1n2 = sin i/sin r
(Since , for small angle sin theta =theta)
u2/u1=i/r
n1*i=n2*r -------- (1)
In triangle OAC
i = Alpha +gamma ------(2)
And in triangle IAC
r+ beta = gamma
r=gamma - beta --------(3)
Therefore , from eq. (1) ,
u1 (alpha+gamma) = u2 (gamma - beta)
u1*alpha + u1*gamma =u2*gamma -u2*beta
(Since , alpha=tan(alpha) =AM/MO
beta=tan(beta)=AM/MI
gamma=tan(gamma)=AM/MC
)
u1(AM/MO +AM/MC) = u2(AM/MC + AM/MI)
(Since , M is very closed to P)
u1(1/PO+1/PC)=u2(1/PC-1/PI)
(Since PO= -u
PC=+r
PI=+v )
-u1/u + u1/r = u2/r - u2/v
-u1/u + u2/v =u2/r - u1/r
u2/v - u1/u = (u2-u1)/r
Hence , proved!!
Derivation:
The ray diagram required is attached in the image below.
From ΔAOC, i = α + γ
From ΔAIC, γ = r + β ⇒ r = γ - β
By Snell's law, we get,
μ₁ sin i = μ₂ sin r
Since, the angles are small, we get,
μ₁ i = μ₂ r
⇒ μ₁ tan i = μ₂ tan r
On substituting i and r, we get,
μ₁ tan (α + γ) = μ₂ tan (γ - β)
μ₁ (tan α + tan γ) = μ₂ (tan γ - tan β)
μ₁ (AM/PO + AM/MC) = μ₂ (AM/MC - AM/MI)
μ₁ (1/PO + 1/MC) = μ₂ (1/MC - 1/MI)
∵ MC = PC and MI = PI
μ₁/PO + μ₁/PC = μ₂/PC - μ₂/PI
By sign convention, we get,
PO = - u; PC = R; PI = v
⇒ μ₁/- u + μ₁/R = μ₂/R - μ₂/v
μ₂/v - μ₁/u = μ₂/R - μ₁/R
∴ μ₂/v - μ₁/u = (μ₂ - μ₁)/R
Hence deduced.
