Resolving power of a microscope is defined as the reciprocal of its limit of resolution (d) i.e.
RP of microscope = 1/d
where, limit of resolution is equal to the smallest distance between two closest objects whose vivid or clean image can be seen through the
microscope and given by d = \(\frac { \lambda }{ 2\mu sin\theta } \)
Resolving power of microscope = \(\frac { 2\mu sin\theta }{ \lambda } \)
where, \(\lambda \) = wavelength of light used,
\(\theta =\) semivertical angle of the cone formed by object
at objective and \(\mu \)  = refractive index of molecule between object and lens.
(a) Resolving power increases with the increase of \(\mu \)
(b) Resolving power decreases as resolving power \(\propto 1/\lambda \)

Ray diagram of a reflecting type telescope
Advantages :
(i) Reflecting telescopes have high resolving power due to a large aperture of mirrors.
(ii) Due to availability of paraboloidal mirror, the image is free from chromatic and spherical aberration.

f = 40 cm

Polarization.
Two Uses:
Polaroids can be used in sunglasses, window panes, photographic cameras, 3D movie cameras

Direction; Condition: Size of the obstacle sharpness should be comparable to the wavelength of the light falling.
Application: the finite resolution of our eye.

Angular separation, \(\theta={\beta\over D}={\lambda\over d}\)
It does not depend upon D, the distance of separation between slits and screen. Therefore, \(\theta\) remains unaffected.

(i) From a point source, the wavefront is diverging spherical wavefront is diverging spherical wavefront.
(ii) From a distant light source, the wavefront is plane wavefront.

A wavefront is the locus of all such particles of the medium, which are vibrating in the same phase.

-15 cm, from 1/f = 1/f₁ + 1/f₂ 

Consider any point Q on the incident wavefront.

Suppose when disturbance from point P on incident wavefront reaches point P' on the refracted wavefront, the disturbance from point Q reaches point Q' on the refracting surface XY. Since A'Q'P' represents the refracted wavefront. the time taken by light to travel from a point on' incident wavefront to the corresponding point on refracted wavefront should always be the same. Now, time taken by light to go from Q to Q' will be
\(t=\frac { QK }{ c } +\frac { KQ' }{ v } \quad \quad .....(i)\)
(where, c and v are the velocities of light in two mediums)
In right angle \(\triangle AQK,\ \angle QAK=i\)
QK = AK sin i .... (ii)
In right angled \(\triangle P'Q'K, \ \angle Q'P'K=r\)
KQ' = KP' sin r   .......(iii)
Substituting Eqs. (ii) and (iii) in Eq. (i), we get
\(t=\frac { AK\sin { i } }{ c } +\frac { KP\sin { r } }{ v } \)
\(t=\frac { AK\sin { i } }{ c } +\frac { (AP'-AK)\sin { r } }{ v } \)
\(or\quad t=\frac { AP' }{ v } \sin { r } +\left( \frac { \sin { i } }{ c } -\frac { \sin { r } }{ v } \right) AK\quad \quad .....(iv)\)  
The rays from different points on the incident wavefront will take the same time to reach the corresponding points on the refracted wavefront, i.e., given by Eq. (iv) is independent of AK. It will happen so, if
\(\frac { \sin { i } }{ c } -\frac { \sin { r } }{ v } =0\)
\(\Rightarrow \quad \frac { \sin { i } }{ \sin { r } } =\frac { c }{ v } \)
However, \(\frac { c }{ v } =n\)
This is the Snell's law for refraction of light.

In an unpolarised light, the vibrations of electric field vector are in every plane perpendicular to the direction of propagation of light.
When unpolarised light is incident on the boundary between two transparent media, the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. 1 Brewster angle: 
\(\mu =tan{ i }_{ B }\) .

For convex lens of focal length 10 cm,
f =+ 10cm, u = -3 cm
Using lens formula,
\( \frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u }\)  
\(\Rightarrow \ \frac { 1 }{ 10 } =\frac { 1 }{ v } -\frac { 1 }{ \left( -30 \right) } \Rightarrow v=15cm\)
The image formed by first lens acts as a virtual object for plano-concave lens.
For plano-concave lens,
 \(u=+10cm,\quad f=-10cm,v=?\) 
Using lens formula,
 \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \) 
\(\Rightarrow \quad \frac { 1 }{ v } =0\)  
\( \Rightarrow \quad v=\infty \)
The refracted ray becomes parallel to principal axis for convex lens of focal length 30 cm.
 \(u=-\infty ,v=?,f=30cm\) 
 \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \) 
\(\Rightarrow \ \frac { 1 }{ 30 } =\frac { 1 }{ v } -\frac { 1 }{ \left( -\infty \right) } \ \Rightarrow \ v=30cm\) 
So, final image is formed at a distance of 30 cm from second convex lens on the other side of u.   

\(Here,\ \lambda =600nm=6\times { 10 }^{ -7 }m,\)
\(a=3mm=3\times { 10 }^{ -3 }m\)
\( D=3m;\left( 2x \right) =?\)
\(2x=\frac { 2\lambda D }{ a } =\frac { 2\times 6\times { 10 }^{ -7 }\times 3 }{ 3\times { 10 }^{ -3 } }\)
\(=12\times { 10 }^{ -4 }m=1.2mm\)

Here,\(Here,{ R }_{ 1 }=10cm,{ R }_{ 2 }=-15cm,f=12cm\)
\(\mu =?\)
From lens makers formula
\(\frac { 1 }{ f } =(\mu -1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ 12 } =(\mu -1)\left( \frac { 1 }{ 10 } +\frac { 1 }{ 15 } \right) =(\mu -1)\frac { 1 }{ 6 }\)
\( \\ \mu -1=\frac { 6 }{ 12 } =\frac { 1 }{ 2 } ,\mu =\frac { 1 }{ 2 } +1=\frac { 3 }{ 2 } =1.5\)

(a) The ray diagram, showing image formation by a compound microscope, is given below:

Linear magnification due to the objective = \(=\frac { h^{ ' } }{ h } =\frac { L }{ f_{ 0 } } \)
\(\left( \therefore tan\beta =\frac { h }{ f_{ 0 } } =\frac { h^{ ' } }{ L } \right) \)
Here, L = tube length = distance between the second focal point of the objective and the first focal point of the eyepiece.
When the final image is formed at infinity, the angular magnification due to the eyepiece equals
\(\frac { D }{ f_{ e } } \) (D = least distance of distinct vision)
\(\therefore \) Total magnification when the final image is formed
at infinity = \(\left( \frac { L }{ f_{ 0 } } ,\frac { D }{ f_{ e } } \right) \)
(b) Resolving power increases when the focal length of the objective is decreased.
(i) This is because the minimum separation \(d_{ min }\left( =\frac { 1.22f\lambda }{ D } \right) \) decreases when f is decreased
(ii) Resolving power decreases when the wavelength of light is increased.
This is because the minimum separation,  \(d_{ min }\left( =\frac { 1.22f\lambda }{ d } \right) \) increases when \(\lambda \) is increased.

(a) The light rays from the two (coherent) slits, reaching a point 'P' on the screen, have a path
difference (S₂P - S₁P). The point 'P' would, therefore be a 
(i) Point of maxima (bright fringe), If
S₂P - S1P = n\(\lambda\) .
(ii) Point of minima (dark fringle), If
S₂P- S₁P = (2n + 1) \(\lambda\)/2

We have (S₂P)²- (S₁P)²
\(=\left\{D^2-\left(x+{d\over2}\right)^2\right\}-\left\{D^2+\left(x-{d\over2}\right)^2\right\}\)
= 2 xd
\(S_2P-S_1P={2xd\over S_3P+S_1P}={2xd\over 2D}={xd\over D}\)
We have maxima at point, where
\({xd\over D}=n\lambda\)
and minima at points where
\({xd\over D}=\left(2n+1\over 2\right)\lambda\)
Now, fringe width \(\beta\)  = separation between two successive maxima (or two successaive minima)
= x_n - x_n-1
\(\beta={\lambda D\over d}\)
(b) We have \({I_{max}\over {I_{min}}}={(a_1+a_2)^2\over (a_1-a_2)^2}={25\over 9}\)
\(\therefore\ \ {a_1+a_2\over a_1-a_2}={5\over 3}\Rightarrow{a_1\over a_2}={4\over 1}\)
\(\therefore\ {W_1\over W_2}={I_1\over I_2}={(a_1)^2\over (a_2)^2}={16\over 1}\)  

For refraction at the first surface
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For the second surface, I₁ acts as a virtual object (located in the denser medium) whose final real image is formed in the rarer medium at I.
so for refraction at this surface, we have
\(\frac { { n }_{ 2 } }{ v } -\frac { { n }_{ 1 } }{ { v }_{ 1 } } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 2 } } \)
From the above two equation,  \(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
The point, where image of an object, located at infinity is formed, is called the focus F, of the lens and the distance f gives its focal length.
So for \(u=\infty ,v=+f\)
\(\Rightarrow\)  \(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
(b)

\(\triangle\) ABP is similar to \(\triangle\)A'B'P
So    \(\frac { A'B' }{ AB } =\frac { B'P }{ BP } \)
Nor A'B' = I, AB = O, B'P = + v and BP = - u
So magnification \(m=\frac { I }{ O } =-\frac { v }{ u } \)