(i) Consider a thin lens made up of a medium of refractive index n₂ placed in a medium of refractive. index n₁ on its left and medium of refractive index n₃ on its right. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.
(ii) A paraxial ray from O forms image at I' after refraction at surface (1). Before that it is again refracted by the surface (2) due to which final image is formed at I.

For the refracting surface (1), the light goes from n₁ to n₂.
\(\frac{\mathrm{n}_{2}}{\mathrm{v}^{\prime}}-\frac{\mathrm{n}_{1}}{\mathrm{u}}=\frac{\mathrm{n}_{2}-\mathrm{n}_{1}}{\mathrm{R}_{1}}\) ...(1)
For the refracting surface (2), the light goes from n₂ to n₃.
Then, \(\frac{\mathrm{n}_{3}}{\mathrm{v}}-\frac{\mathrm{n}_{2}}{\mathrm{v}^{\prime}}=\frac{\mathrm{n}_{3}-\mathrm{n}_{2}}{\mathrm{R}_{2}}\) ......(2) 
Add equations (1) and (2)
\(\frac{n_{3}}{v}-\frac{n_{1}}{u}=\frac{n_{2}-n_{1}}{R_{1}}+\frac{n_{3}-n_{2}}{R_{2}}\)  ....(3)  

\(n =\frac{1}{\sin c} \)
\(=\frac{1}{\sin 45^{0}}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}=1.414 \)
n_R = 1.39
n_G = 1.44
n = 1.47
Here n_R < n
Since n_R < n, Red colour only will emerge out of prism.
Since \(n_{G}>n \ \& \ n_{B}>n\) Blue and green undergo total internal reflection.

Given, n = 1.5; R₁ = 30 cm; R₂ = –30 cm;
(a) Let us find f_l and f_m separately.
Using lens maker’s formula we can find f_l .
\(\frac{1}{f_{l}}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) 
Substituting the values, \(\frac{1}{f_{l}}=(1.5-1)\left(\frac{1}{30}-\frac{1}{-30}\right)\) 
\(\frac{1}{f_{l}}=(0.5)\left(\frac{1}{30}+\frac{1}{-30}\right)=\left(\frac{1}{2}\right)\left(\frac{2}{30}\right)=\frac{1}{30 \mathrm{~cm}}\) 
f _l = 30 cm = 0.3 m
Focal length of mirror is, \(f_{m}=\frac{R_{2}}{2}\) 
Substituting the values, \(f_{m}=\frac{-30}{2}=-15 \mathrm{~cm}\) 
f _m = 15 cm = –0.15 m
Now the focal length of the slivered lens is,
\(\frac{1}{-f}=\frac{2}{f_{l}}+\frac{1}{-f_{m}}=\frac{2}{30}+\frac{1}{15}=\frac{2}{15}=\frac{1}{7.5}\) 
f = – 7.5 cm = –0.075 m
The silvered mirror behaves as a concave mirror with its focal length on left side.
To find the power of the silvered lens,
P = 2P_l + P_m 
\(P=\frac{2}{f_{l}}+\frac{1}{-f_{m}}=\frac{2}{0.3}+\frac{1}{-(-0.15)}\) 
\(=\frac{2}{0.3}+\frac{1}{0.15}=\frac{4}{0.3}=13.33 D\) 
As the power is positive it is a converging system.
[Note: Here, we come across a silvered lens which has negative focal length and positive power. Which implies that the focal length is to the left and the system is a converging one. Such situations are possible in silvered lenses because a silvered lens is basically a modified mirror.]
(b) Writing the mirror formula,
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\) 
Here, both u and v are same (v = u) as the image coincides with the object.
\(\frac{1}{-7.5 \mathrm{~cm}}=\frac{1}{u}+\frac{1}{u}=\frac{2}{u} ; \mathrm{u}=-2 \times 7.5 \mathrm{~cm}\) 
u = –15 cm = –0.15 m
The object is to be placed 15 cm to the left of the silvered lens.

The full size of the image will be obtained. But the intensity of image will be reduced. This is because the number of rays of light refracted through different parts of the lens will be reduced.
.

\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) or \(\cfrac { 1 }{ v } =\cfrac { 1 }{ f } +\cfrac { 1 }{ u } \) 
convex lens : f > 0 ; u < 0 (object on the left)
For 0 < I u I \(\cfrac { 1 }{ v } =\cfrac { 1 }{ f } -\cfrac { 1 }{ |u| } <0\) i.e. u < 0 (image on left; virtual)
Also for this case, \(\cfrac { 1 }{ \left| v \right| } <\cfrac { 1 }{ |u| } .i.e.\left| u \right| <\left| v \right| \) (image enlarged)

From the lens equation
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ -f } =\cfrac { 1 }{ f } \) 
or \(\cfrac { 1 }{ v } =0\)
∴v = ∞
So, the image is formed at infinity.

(i) A lens is formed by a transparent material bounded between two spherical surfaces or one plane and another spherical surface.
(ii) In a thin lens, the distance between the surfaces is very small. If there are two spherical surfaces, then there will be two centers of curvature C₁ and C₂ and correspondingly two radii of curvature R₁ and R₂.
(iii) A plane surface has its center of curvature C at infinity and its radius of curvature R is infinity (R = ∞).

Refractive index of a transparent medium is defined as the ratio of speed of light in vacuum (or air) to the speed of light in that medium refractive index n of a medium.
= \(\cfrac { speed\ of lightin\ vacuum\left( c \right) }{ speed\ of \ light\ inmedium\left( v \right) } \)

(i) When light rays incident on a spherical mirror, after reflection, the rays converge at a point on the principal axis for concave mirror or appear to diverse from a point for convex Mirror.
(ii) This point is called the focus or focal point (F) of the mirror.

(i) The line joining the pole and the centre of curyature is called the principal axis of the mirror.
(ii) The light ray travelling along the principal axis towards the mirror after reflection travels back along the same principal axis. It is also called optical axis.

(i) A microscope is used to see the details of the object under observation.'
(ii) The ability of microscope depends not only on magnifying the object but also on resolving two points on the object separated by a small distance d .. nun
(iii) Smaller the value of d_min better will be the resolving power of the microscope.
(iv) The radius of central maxima is already derived as equation (1),
\(N=\cfrac { 1 }{ a+b } \) 
\({ r }_{ 0 }=\cfrac { 1.22\lambda f }{ a } \) 
(v) In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is d_min, then the magnification m is
\(m=\cfrac { { r }_{ 0 } }{ d_{ min } } \) 
\({ d }_{ min }=\cfrac { { 1.22 }\lambda v }{ m } =\cfrac { 1.22\lambda v }{ a\left( \cfrac { v }{ u } \right) } =\cfrac { 1.22 }{ \lambda v } =\cfrac { 1.22\lambda u }{ a } \)  \(\left[ \therefore m=\cfrac { v }{ u } \right] \) 
\({ d }_{ min }=\cfrac { 1.22\lambda f }{ a } \) \(\left[ \therefore u=f \right] \) 
On the object side,
\(2tan\beta =2sin\beta =\cfrac { a }{ f } \) \(\left[ \therefore a=f2sin\beta \right] \) 
\({ d }_{ min }=\cfrac { 1.22\lambda }{ 2sin\beta } \)
(vi) To reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.

(i) Light is allowed to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously.
(ii) As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams.
(iii) They will be differences in either in-phase or at constant phase. Instruments like Michelson's interferometer, Fabray- Perrot etalon work on this principle.

According to law of reflection,
(i) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(ii) The angle of incidence i is equal to the angle of reflection r.
i = r