`(i)` Device the mathematical `n` between refractive indices `n_(1)` and `n_(2)` of two radii and radius of curvature `R` for refraction at a convex shperical surface. Consider the object to be a point since lying on the principal axis in rarer medium of refractive index `n_(1)` and a real image formed in the denser medium of refractive index `n_(2)`. Hence, derive lens marker's formula.
`(ii)` Light from a point source in air falls on a convex spherical glass surface of refractive index `1.5` and radius of curvature `20cm`. The distance of light source from the glass surface is `100cm`. At what position is the image formed ?

Explanation

Let `i` be the angle of incidence of ray `OA` and `r` the anlge of refraction in the densor medium i.e, `/_OAN=i` and `/_CAI=r`. Let `/_AOP=a /_AIP=B` and `/_ACP=b` and `/_CP ACP=y`
In triangle `OAC` `i=y+a`
In triangle `AIC`, `y=b+t` or `r=y-B`
From Snell's law
If point `A` is very close to `P`, the angles `i,r,a,b` and `c` will be very small. Therefore
`sin i=i` and `sinr=r`
From equation,
`(i)/(r )=(n_(2))/(n_(1))`
Sunstituting values of `i` and `r` , we get
`(gamma+alpha)/(gamma-beta)=(n_(2))/(n_(1))` or `n_(1)(gamma+alpha)/(gamma-beta)=(n_(2))/(n_(1))` or `n_(1)=(gamma+alpha)=n_(2)(gamma-beta)`
Let `h` be the height of perpendicular drawn from `A` on principal axis i.e. `AM=h`. As `a,b` and `y` are very small angles.
`:. tanalpha=alpha`, `tanbeta=beta` and `tangamma=gamma`
Substituting these values,
`n_(1)(tangamma+tan alpha)=n_(2)(tangamma-tanbeta)`
As point A is very close to point `P`, Point `M` is coincident with `P`.
From figure `tan alpha=(AM)/(OM)=(h)/(OP)impliestanbeta=(AM)/(MI)=(h)/(PI)`
`tangamma=(AM)/(MC)=(h)/(PC)`
Substituting these values , we get
`n_(1)((h)/(PC)+(h)/(OP))=n_(2)((h)/(PC)+(h)/(PI))` or `n_(1)((1)/(PC)+(1)/(OP))=n_(2)((1)/(PC)+(1)/(PI))`
If the distances of object `O`, image `I`, centre of curvature `C` from the pole be `u,v` and `R` respectively, then by sign convention `PO` is negative while `PC` and `PI` are positive. Thus,
`u=-OP`, `v=+PI`, `R=+PC`
Substituting these values in `(vi)`, we get
`n_(1)((1)/(R)-(1)/(u))=n_(2)((1)/(R )+(1)/(v))` or `(n_(1))/(R )-(n_(1))/(R )-(n_(1))/(u)=(n_(2))/(R )-(n_(2))/(u)`
`(n_(2))/(v)-(n_(1))/(u)=(n_(2)-n_(1))/(R )`