Answers

2015-02-24T19:02:20+05:30
Let SPS' be a convex mirror surface separating a rarer medium of absolute refractive index n₁ from a denser medium of refractive index n₂.Let P be the centre of curvature of the surface .PC is the radius of curvature R of the surface.Let a point-object O placed on the principal axis PC produced backwards.An incident ray of light OA,after reflection at the point A on the surface,bends towards the normal CAN and goes along AI in the denser medium.Another incident ray OP falls on the refracting surface normally and goes un-deviated into the denser medium.The 2 refracted rays meet at the principal axis at I which is the real image of O.
Suppose the angle of incidence OAM=i,the angle of refraction IAC=r ,Po=-u,PI=+v and PC=+R.According to the coordinate geometry sign convention,u is -ve while v and R is +ve.Let α,β,γ b the angle made by OA,IA,Ca respect. with proncipal axis.Let h be the height of the normal AM dropped from A on the principal axis.Now,in ΔCOA,
i=α+γ
γ=r+β........[in CIA]
or r=γ-β
by snell's law:
sini/sinr=n2/n1
or n₁ sin i= n₂ sin r
or n₁i=n₂r......................[since the spherical mirror is of small aperture]
subtitute values
n₁(α+β)=n₂(γ-β)
or,n₁+α+n₂+β=(n₂-n₁)
therefore,α=tanα =AM/Mo=Am/Po=h/-u
β=tanβ=Am/Mi=AM/PC=h/v
γ=tanγ=AM/MC=AM/PC=h/r

n₂(h/v)+n₁(h/-u)=(n₂-n₁)h/r
orn₂/v-n₁/u=(n-n₁)/R
therefore the required eq. is
n/v-i/u=(n-1)/R
2 5 2