1. Refractive index of medium 2 is greater than medium 1.

2. Refraction at a spherical surface

Consider a convex surface XY, which separates two media having refractive indices n_{1} and n_{2} . Let C be the centre of curvature and P be the pole. Let an object is placed at ‘O’, at a distance ‘u’ from the pole.

I is the real image of the object at a distance V from the surface. OA is the incident ray at angle ‘i’ and Al is the refracted ray at an angle ‘r’. OP is the ray incident normally. So it passes without any deviation.

From snell’s law,

\(\frac{sin\,i}{sin\,r}=\frac{n_2}{n_1}\)

If ‘i’ and ‘r’ are small, then sin i » i and sin r » r.

i.e.,\(\frac{i}{r}=\frac{n_2}{n_1}\)

n_{1}i = n_{2}r ……….(1)

From the Δ OAC, exterior angle = sum of the interior opposite angles

i-e., i = α + θ ………(2)

Similarly, from Δ IAC,

α = r + β

r = α – β……..(3)

Substituting the values of eq(2) and eq(3)in eqn. (1) we get,

n_{1}(α + θ) = n_{2}(α – β)

n_{1}α + n_{1}β = n_{2}α – n_{2}β

n_{1}θ + n_{2}β = n_{2}α – n_{1}α

n_{1}θ + n_{2}β = (n_{2} – n_{1})α ……….(4)

From OAP, we can write,

θ = \(\frac{AP}{OP} [angle = \frac{arc}{radius}]\)

From IAP, β = \(\frac{AP}{PI}\),From CAP, a = \(\frac{AP}{PC}\)

Substituting θ, β, and α in equation (4) we get,

n_{1}\(\frac{AP}{OP}+n_2\frac{AP}{PI}=(n_2-n_1)\frac{AP}{PC}\)

\(\frac{n_1}{OP}+\frac{n_2}{PI}=\frac{(n_2-n_1)}{PC}\)

According to New Cartesian sign convection, we can write,

OP = -u, PI = +v and PC = R

Substituting these values, we get

\(\frac{-n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}\)

\(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\)

**Case -1:**

If the first medium is air, n_{1} = 1, and n_{2} = n,

\(\frac{n}{v}-\frac{1}{u}=\frac{n-1}{R}\)

3. Consider a thin lens of refractive index n_{2} formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n_{1} . Let an object ‘O’ is placed in the medium of refractive index n_{1} Hence the incident ray OM is in the medium of refractive index n_{1} and the refracted ray MN is in the medium of refractive index n_{1} .

The spherical surface ABC (radius of curvature R_{1}) forms the image at I_{1} Let ‘u’ be the object distance and ‘v_{1}’ be the image distance.

Then we can write,

\(\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}\) ....(1)

This image I_{1} will act as the virtual object for the surface ADC and forms the image at v.

Then we can write,

\(\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}\) .....(2)

Adding eq(1) and eq(2) we get

\(\frac{n_2}{v_1}-\frac{n_1}{u}+\frac{n_1}{v}\)- \(\frac{n_2}{v_1}\) = \(\frac{n_2-n_1}{R_1}+\frac{n_1-n_2}{R_2}\)

\(\frac{n_1}{v}-\frac{n_1}{u}\) = (n_{2 }- n_{1})\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

Dividing throughout by n_{1} , we get

\(\frac{1}{v}-\frac{1}{u}\) = \(\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

if the lens is kept in air, \(\frac{n_2}{n_1}=n\)

So the above equation can be written as,

\(\frac{1}{v}-\frac{1}{u}\) = (n-1)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) .....(4)

From the definition of the lens, we can take, when

U = ∞, f = v

Substituting these values in the eq (3), we get

∴ \(\frac{1}{f}-\frac{1}{\infty}\) = (n - 1)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

This is lens maker’s formula

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) .....(5)

For convex lens,

f = +ve, R_{1} = +ve, R_{2} = – ve

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)

For concave lens,

f = -ve, R_{1} = -ve, R_{2} = +ve

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

Lens formula From eq(4),

\(\frac{1}{v}-\frac{1}{u}=(n-1)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

From eq (5)

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

4. u = -8cm, f = +4cm

m = \(\frac{f}{f+u}=\frac{4}{4+-8}\)

m = -1.