\(\cfrac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)

There will be an increases in the focal length of the convex lens. This is because the refractive index of glass with respect to water is less than the refractive index of glass with respect to air.

(i) Focal length increases with increase of wavelength.\(\cfrac { 1 }{ f } \left( \cfrac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } -1 \right) \) as wavelength increases  \(\cfrac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } \)  decreases hence focal length increases.
(ii) As μ_l increase focal length increases.\(\cfrac { 1 }{ f } \left( \cfrac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } -2 \right) \cfrac { 2 }{ R } \)

(i) Fringe width of interference fringes, is given by \(\beta =\cfrac { D\lambda }{ d } \propto \cfrac { 1 }{ d } \) . If the sources are far apart, d is large; so fringe width (β) will be so small that the fringes are not resolved and they do not appear separate.
(ii) That is why the interference pattern is not detected for large separation of coherent sources.

The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

(i) The incident light is assumed to be monochromatic (single colour).
(ii) The incident ray of light is very dose to the principal axis (paraxial rays).

(i) To study the spectra of different sources of light.
(ii) To measure the refractive indices of materials due to determinant of angle of prism (A) and angle of minimum deviation (D).
(iii) To find the refractive indices of liquids.
(iv) To find the wavelength of light using grating.

The tangent of the polarising angle for transparent medium is equal to its refractive index.
tan i_p = n

This is the most commonly used method for producing two coherent sources. If two points are chosen on the wavefront by using a double slit, the two points will act as coherent sources.

A wavefront is the locus of points which are in the same state or phase of vibration.

Quantum theory of light:
Quantum theory states that light waves consist of small packets of energy called photons. The energy associated with each photon is E = hv, Where 'h' is Planck's constant (h = 6.625 x 10^-34 J s) and v is frequency of electromagnetic radiation.

(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 (i) and (ii) 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 (i) forms image at I'.
(iii) Before it does so, it is again refracted by the surface (ii). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given by the equation is,
\(\cfrac { { n }_{ 2 } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { \quad n }_{ 2 }-{ { n }_{ 1 } } \right) }{ R } \)   ....(i)
(v) For the refracting surface (1), the light goes from n₁ to n₂
 \(\cfrac { { { n }_{ 2 } } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( n_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1} } \)   ......(ii)

(vi) For the refracting surface (ii), the light goes from medium n₂ to n₁
 \(\cfrac { { { n }_{ 1 } } }{ v' } -\cfrac { { n }_{ 2 } }{ v' } =\cfrac { \left( n_{ 1 }-{ n }_{ 2 } \right) }{ { R }_{ 2 } } \)   ....(iii)
(vii) Adding the above two equations (ii) and (iii)
\(\cfrac { { { n }_{ 1 } } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)
On further simplifying and rearranging
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)    ....(iv) 
If the refractive of the lens is n₂ and is placed in air, then n₂= n and n₁= 1. So the equation (iv) becomes,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{u } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ...(v)
According to lens makers formula
\(\cfrac { 1 }{ f } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ...(vi) 
Comparing the two equations (v) and (vi), we find
 \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u }=\cfrac { 1 }{ f }\)          .....(vii)
This is called lens equation

An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient's body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body.

Yes, It is possible for two lens to produce zero power.
Explanation:
when the two lens (one is concave & another one is convex lens) are combined together, the focal length of the combination of the two lenses is F.
Then
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)
\(\text { if } f_{1}=f_{2}=f \)
\(f_{1}=f(\text { convex }) \)
\(f_{2}=-f(\text { concave) } \)
\(P=\frac{1}{F}=\frac{1}{f}-\frac{1}{f}=0 . \ P \rightarrow \text { power } \)