\(\cfrac { 1 }{ F } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) 
\(\cfrac { 1 }{ 50 } =\cfrac { 1 }{ 20 } +\cfrac { 1 }{ { { f }_{ 2 } } } \quad f=-\cfrac { 100 }{ 3 } cm\) 
\({ P }_{ 2 }=\cfrac { 100 }{ -\frac { 100 }{ 3 } } \) 
D = -3D.

Net power, P = P₁ + P₂
or P = -4D + 2D = -2D
Focal length,\(f=\cfrac { 1 }{ P } =\cfrac { 1 }{ -2 } =-50cm\) 

Formula:
Refractive index = \(A+\cfrac { B }{ { \lambda }^{ 2 } } \) 
where λ is the wavelength.
Power of a lens \(P=\cfrac { 1 }{ f } =\left( { n }_{ g }-1 \right) \left( \cfrac { 1 }{ { { R }_{ 1 } } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 

Clearly, the power of a lens (n_g - 1). This implies that the power of a lens decreases with the increase in wavelength \(\left( P\infty \cfrac { 1 }{ { \lambda }^{ 2 } } nearly \right) \) The plot is shown in the figure alongside.

The focal length of the lens (refractive index μ₁) in a liquid of refractive index μ₂ is
Formula:
\({ f }_{ 1 }=\cfrac { { \mu }_{ 1 }-1 }{ \frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } } \times { f }_{ a }\) 
Given :
\({ \mu }_{ 2 }>{ \mu }_{ 1 },i.e,\cfrac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } <1\) 
So,\({ \quad f }_{ 1 }=\cfrac { { \mu }_{ 1 }-1 }{ 1-\frac { { \mu }_{ 1 } }{ { \mu }_{ 2 } } } { f }_{ a }\)
So the focal length of the lens in the liquid will be of the opposite sign of the focal length of the lens in air, i.e., the nature of the lens will change. Hence, the lens would now behave like a diverging (concave) lens.

For face AB, ∠i = 0°,∠r = ·0° the ray will pass through AB undeflected 
Now, at face AC

Formula.
Here,\({ \quad i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ \mu } \right) \) 
= \({ sin }^{ -1 }\left( \cfrac { 2 }{ 3 } \right) ={ sin }^{ -1 }\left( 0.66 \right) \) 
∠i on face AC is 30^o which is less than ∠i_e.
Hence, the ray get refracted.
And, applying Snell's law at face AC
\({ sin30 }^{ o }\times \cfrac { 3 }{ 2 } =sinr\times 1\) 
\(\Rightarrow sinr=\cfrac { 1 }{ 2 } \times \cfrac { 3 }{ 2 } \) 
\(\Rightarrow r={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) ={ sin }^{ -1 }(0.75)\)
And, dearly r > i, as ray passes from denser to rarer medium.

Here A = 60^o,\({ \delta }_{ m }=30^{ o }\)
We know that
Formula:
\(\mu =\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \frac { A }{ 2 } \right) } \) 
= \(\cfrac { sin\left( \frac { { 60 }^{ o }+{ 30 }^{ 0 } }{ 2 } \right) }{ sin\left( \cfrac { { 60 }^{ o } }{ 2 } \right) } \) 
= \(\cfrac { sin{ 45 }^{ o } }{ sin{ 30 }^{ o } } =\sqrt { 2 } \) 
Also  \(\mu =\cfrac { c }{ v } \Rightarrow \cfrac { 3\times { 10 }^{ 8 } }{ \sqrt { 2 } } m/s\) 
= 2.122 x 10⁸ m/s
(ii) 

At face AC, let the angle of incidence be r₂. For grazing ray, e = 90°
\(\mu =\cfrac { 1 }{ sin{ r }_{ 2 } } \) 
\(\Rightarrow { r }_{ 2 }={ sin }^{ -1 }\left( \cfrac { 1 }{ \sqrt { 2 } } \right) ={ 45 }^{ o }\) 
Let angle of refraction at face AB be r₁ 
Now r₁ + r₂ = A
ஃr₁ = A - r₂ = 60^o- 45^o = 15^o
Let Angle of incidence at this face be i
\(\mu =\cfrac { sini }{ sin{ r }_{ 1 } } \Rightarrow \sqrt { 2 } =\cfrac { sini }{ sin15^{ o } } \) 
\(\therefore i={ sin }^{ -1 }\left( \sqrt { 2 } .sin{ 15 }^{ o } \right) \)

For the convex lens 
Formula :\(\cfrac { 1 }{ { f }_{ 1 } } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ +30 } =\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } \) 
\(\cfrac { 1 }{ { v }_{ 1 } } =\cfrac { 1 }{ 30 } -\cfrac { 1 }{ 40 } =\cfrac { 1 }{ 120 } \) 
v_I = 120 cm a real image is formed.
On introducing a concave lens
f₂ = 50cm and u₂ = 120 - 20 = +100 cm from the concave lens
\(\cfrac { 1 }{ { f }_{ 2 } } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ { u }_{ 2 } } \) 
\(\cfrac { 1 }{ -50 } =\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ +100 } \) 
\(\therefore \cfrac { 1 }{ { v }_{ 2 } } =\cfrac { 1 }{ 580 } +\cfrac { 1 }{ 100 } =\cfrac { 1 }{ 100 } \)
v₂ = -100cm

(i) At a minimum deviation \(\mu =\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \cfrac { A }{ 2 } \right) } \) 
 Given \({ \delta }_{ m }=A\)
\(\mu =\cfrac { sinA }{ sin\cfrac { A }{ 2 } } =\cfrac { 2sin\cfrac { A }{ 2 } cos\cfrac { A }{ 2 } }{ sin\cfrac { A }{ 2 } } \) 
= \(2cos\cfrac { A }{ 2 } \) 
\(\therefore cos\cfrac { A }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \cfrac { A }{ 2 } =30\) A = 6^o 
(iii) \(\mu =\sqrt { 3 } \cfrac { 1 }{ { sini }_{ c } } \Rightarrow { sini }_{ c }=\cfrac { 1 }{ \sqrt { 3 } } \) 
ஃ Angle of incidence > i_c
Total internal reflection takes place.

R = -20 cm and M = -2
Focal length \(f=\cfrac { R }{ 2 } =-10cm\) 
Magnification \(M=\cfrac { -v }{ u } =-2\) 
\(\therefore v=2u\) 
Using mirror formula
\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \Rightarrow \cfrac { 1 }{ 2u } +\cfrac { 1 }{ u } =-\cfrac { 1 }{ 10 } \) 
\(\cfrac { 3 }{ 2u } =-\cfrac { 1 }{ 10 } \Rightarrow u=-15\) 
v = 2 (-15)  = -30 cm
(iii) \(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
Using sign convention for convex mirror we get
f > 0, U < 0
ஃFrom the formula:
\(\cfrac { 1 }{ v } =\cfrac { 1 }{ f } -\cfrac { 1 }{ u } \) 
As f is positive and u is negative, v is always positive, hence the image is always virtual.

(i) If refraction occurs at the first surface
\(\cfrac { { \mu }_{ 1 } }{ { v }_{ 1 } } -\cfrac { 1 }{ u } =\left( \cfrac { { \mu }_{ 1 }-1 }{ R } \right) \) 
If refraction occurs at the second surface, and the image of the first surface acts as an object.
\(\cfrac { { \mu }_{ 2 } }{ v } -\cfrac { { u }_{ 1 } }{ { v }_{ 1 } } =\cfrac { { \mu }_{ 2 }-{ \mu }_{ 1 } }{ -R } \) 
On adding equations (1) and (2), we get
\(\cfrac { { \mu }_{ 2 } }{ v } -\cfrac { 1 }{ u } =\cfrac { 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1 }{ R } \) 
If rays are coming from infinity, i.e., u = -∞ then v = f
\(\cfrac { { \mu }_{ 2 } }{ f } +\cfrac { 1 }{ \infty } =\cfrac { { 2\mu }_{ 1 }-{ \mu }_{ 2 } }{ R } \) 
\(f=\cfrac { { \mu }_{ 2 }R }{ { 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1 } } \) 
(ii) If the combination behaves as a diverging system then f <: O. This is possible only when

\(\Rightarrow 2{ \mu }_{ 1 }-{ \mu }_{ 2 }-1<0\) 
\(2{ \mu }_{ 1 }-{ \mu }_{ 2 }+1\)
\(\Rightarrow { \mu }_{ 1 }<\cfrac { \left( { { \mu }_{ 2 }+1 } \right) }{ 2 } \)
(iii) If the combination behaves as a converging lens then> 0. It is possible only when

\(\Rightarrow { 2\mu }_{ 1 }-{ \mu }_{ 2 }-1>0\) 
\(\Rightarrow 2{ \mu }_{ 1 }->{ \mu }_{ 2 }+1\) 
\({ \mu }_{ 1 }>\cfrac { \left( { \mu }_{ 2 }+1 \right) }{ 2 } \)
The nature of the image formed is real.

Formula.
\(\mu =\cfrac { sini }{ sinr } =\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } \) 
\(\cfrac { { \mu }_{ A } }{ { \mu }_{ B } } =\cfrac { \frac { sini }{ { sinr }_{ A } } }{ \frac { sini }{ { sinr }_{ B } } } =\cfrac { { sinr }_{ B } }{ { sinr }_{ A } } =\cfrac { \frac { { v }_{ 1 } }{ { v }_{ 2 } } }{ \frac { { v }_{ 1 } }{ { v }_{ B } } } \) 
\(\cfrac { { sinr }_{ B } }{ { sinr }_{ A } } =\cfrac { { v }_{ B } }{ { v }_{ A } } \) 
\(r_{\mathrm{A}}<r_{\mathrm{B}} \sin r_{\mathrm{A}}<\sin r_{\mathrm{B}} \Rightarrow v_{\mathrm{A}}<v_{\mathrm{B}}\)
Speed of light in A is lesser.

Formula:
Refractive index,\(\mu =\cfrac { sini }{ sinr } \) 
\(\sqrt { 3 } =\cfrac { sin{ 60 }^{ o } }{ sinr } \) 
\(sinr=\cfrac { \sqrt { 3 } }{ 2 } \times \cfrac { 1 }{ \sqrt { 3 } } =\cfrac { 1 }{ 2 } \) 
sin r = sin30^o
⇒ Angle of refraction = 30^o

\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \cfrac { 2 }{ f } \right) \) 
\(\cfrac { 1 }{ 2 } =\left( \mu -1 \right) \) 
\(\mu =1.5\)

When two thin lenses are separated by a distance d.
(i) Let 0 be a point object on the principal axis of a lens. OA is the incident rayon the lens at a point A at a height h above the optical center.
(ii) The ray is deviated through an angle ઠ and forms the image at me on the principal axis.
(iii) The incident and refracted rays subtend the angles, ∠AOP = α and ∠AIP = β with the principal axis respectively.
In the triangle ∠AOP, the angle of deviation o can be written as,
\(\delta =\alpha +\beta \)   ...(1)
If the height is small as compared to PO and PI, the angles \(\alpha ,\beta \) and 0 are also small. Then,
\(\alpha \approx tan\alpha =\cfrac { PA }{ PO } ;\) and  ...(2)
Then,\(\delta =\cfrac { PA }{ PO } +\cfrac { PA }{ PI } \) 
Here, PA = h, PO = -u and PI = v
\(\delta =\cfrac { h }{ -u } +\cfrac { h }{ v } =h\left( \cfrac { 1 }{ -u } +\cfrac { 1 }{ v } \right) \) 
After rearranging
\(\delta -h\left( \cfrac { 1 }{ v } -\cfrac { 1 }{ u } \right) =\cfrac { h }{ f } \) 
\(\delta =\cfrac { h }{ f } \) 
(iv) The above equation tells that the angle of deviation is the ratio of height to the focal length. Now, the case of two lenses of focal length it and 12 arranged coaxially but separated by a distance d can be considered as shown in the below Figure

\(\delta ={ \delta }_{ 1 }+{ \delta }_{ 2 }\) 
From Equation (5),
\({ \delta }_{ 1 }=\cfrac { { h }_{ 1 } }{ { f }_{ 1 } } ;{ \delta }_{ 2 }=\cfrac { { h }_{ 2 } }{ { f }_{ 2 } } \) and \(\delta =\cfrac { { h }_{ 1 } }{ f } \) 
The equation (6) becomes,
\(\cfrac { { h }_{ 1 } }{ f } +\cfrac { { h }_{ 1 } }{ { f }_{ 1 } } +\cfrac { { h }_{ 2 } }{ { f }_{ 2 } } \) 
From the geometry,
h₂ - h₁ = P₂G - P₂  C = CG
h₂ - h₁ = BG tan ઠ₁≈ BGઠ₁
\({ h }_{ 2- }{ h }_{ 1 }={ h }_{ 1 }d\cfrac { { h }_{ 1 } }{ { h }_{ 2 } } \)
\({ h }_{ 2 }={ h }_{ 1 }d\cfrac { { h }_{ 1 } }{ { h }_{ 2 } } \) 
Substituting the above equation in Equation (8)
 \(\cfrac { 1 }{ f } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } +\cfrac { 1 }{ { f }_{ 1 }{ f }_{ 2 } } \) 
(vi) The above equation could be used to find I the equivalent focal length. To find the position of the equivalent lens, we can further write from the geometry,
\({ PP }_{ 2 }=EG=\cfrac { GC }{ tan\delta } \) 
\({ PP }_{ 2 }=EG=\cfrac { GC }{ tan\delta } \) 
\({ PP }_{ 2 }=EG=\cfrac { GC }{ tan\delta } =\cfrac { { h }_{ 1 }-{ h }_{ 2 } }{ tan\delta } =\cfrac { { h }_{ 1 }-{ h }_{ 2 } }{ \delta } \) 
From equations (7) and (9)
\({ h }_{ 2 }-{ h }_{ 1 }=d\cfrac { { h }_{ 1 } }{ { \quad f }_{ 1 } } \) and \(\delta =\cfrac { { h }_{ 1 } }{ f } \) 
\({ PP }_{ 2 }=\left( d\cfrac { { h }_{ 1 } }{ { f }_{ 1 } } \right) \times \left( \cfrac { f }{ { h }_{ 1 } } \right) \) 
\({ PP }_{ 2 }=\left( d\cfrac { f }{ { f }_{ 1 } } \right) \)

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.

(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (1) forms image at 1'.
(iii) Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given from Equation,
 \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ v} =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R } } \)
For the refracting surface (1), the light goes from n₁ to n₂
\(\cfrac { { n }_{ 2 } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1 } } \)   .....(1)
For the refracting surface (2), the light goes from n₂ to n₁
\(\frac{n_{1}}{v}+\frac{n_{2}}{v^{\prime}}=\frac{\left(n_{1}-n_{2}\right)}{R_{2}}\)    ......(2)
For surface (2) I' acts as virtual object.
Adding the above two equations (1) and (2)
\(\cfrac { { n }_{ 1 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
on further simplifying and rearranging,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ n_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ....(3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \(\infty\), v = f. Then the equation becomes.
\(\cfrac { 1 }{ f } -\cfrac { 1 }{ \infty } =\left( \cfrac { { n }_{ 2 } }{ { { n }_{ 1 } } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ......(4)
If the lens is kept in air, then we can take n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\\ \cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ..(5)
The above equation is called the lens maker's formula.
Significance:
It tells the lens manufacturers what curvature is needed to make a lens of desired focal length with a material of particular refractive index to make a lens of desired focal length. This formula holds good also for a concave lens.