(i) The fringe width interference pattern increase with the decrase in separation between S₁ and S₂ as  \(\beta \alpha \frac { 1 }{ d } \)
(ii) The fring width decreases as wavelength gets reduced, when interferences set up is taken from air to water as, \(\beta \alpha \lambda \)

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.

The focal length of mirror does not change on changing the wavelength of light incident on it. This is because f = R/2, i.e. f depends only on radius of curvature of the mirror.

20; 13.5 cm

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.

Explanation: As per Huygen's Principle, net effect at any point,
= sum total of contribution of all wavelets with proper phase difference.At the central point (O) contribution from each half in SS₁ is in phase with that from the corresponding part in SS₂, Hence, 0 is a maxima  At the point M where SM-S₁M = \(\lambda\) /2 Phase difference between each wavelet from SS₁ and corresponding wavelet from SS₂ = '\(\lambda\) /2. Hence, M would be a minima . All such points (path difference = n\(\lambda\)/2) are also minima . Similarly, all points, for which path difference = (2n + 1)\(\lambda\)/2, are maxima but with decreasing intensity. From the figure Half angular width of central maxima \(\lambda\) /a
\(\therefore\)  Size of central maxima will be reduced to half and intensity of central maxima will be four times if slit is made double the original width.

Magnifying power (in normal adjustments) of a reflecting telescope is the ratio of the focal length of concave reflector and the focal length of the eyepiece.
\(m=\frac { f }{ { f }_{ e } } =\frac { RI2 }{ { f }_{ e } } \)
Limitations of refracting telescope over a reflecting type telescope.
(i) Refracting telescope suffers from chromatic aberration as it uses large sized lenses.
(ii) It is difficult and expensive to make such large sized lenses.

For lens L₁, \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \) 
Given, \(u=-15cm,\ f=+10cm,v=?\) 
\(\therefore \quad \frac { 1 }{ 10 } =\frac { 1 }{ v } +\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ v } =\frac { 1 }{ 10 } -\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ v } =\frac { 1 }{ 30 } \) 
Distance of image from lens L₁, v = 30 cm
For lens \({ L }_{ 3 },\quad \frac { 1 }{ { f }^{ '' } } =\frac { 1 }{ { v }^{ '' } } -\frac { 1 }{ { u }^{ '' } } \) 
Distance of image from lens L₁, v = 10 cm
\(\therefore \ \frac { 1 }{ 10 } =\frac { 1 }{ 10 } -\frac { 1 }{ { u }^{ '' } } \Rightarrow \frac { 1 }{ { u }^{ '' } } =0\Rightarrow { u }^{ '' }=\infty \) 
The refracted rays from lens L₂ becomes parallel to principal axis. It is possible only when image formed by L₁ lies at first focus of L_2, i.e. at a distance of 10 cm from L₂.
\(\therefore \) Separation between L₁ and L₂ = 30 + 10 cm
The distance between L₂ and L₃ may take any value.

\(\left( a \right) 4.4\times { 10 }^{ -7 }m\)
\(\left( b \right) 3.3\times { 10 }^{ -7 }m\)

Light waves, originating from two independent monochromatic sources, will not  have a monochromatic source, will not have a constant phase difference. Therefore, these sources will not be coherent and, therefore, would not produce a sustained interface pattern.
y = y₁ + y₂
= a cos wt + a cos (wt + \(\phi\))
= 2a cos\(\phi\over2\).cos(wt+\(\phi\over2\))
Amplitude of resultant displacement is 2a cos \(\phi\over2\) 
\(\therefore\) Intensity,
I = 4a² cos2 \(\phi\over2\)
A path difference of \(\lambda \) , corresponds to a phase difference of 2\(\pi\)
\(\therefore\) The intensity, K = 4a²  \(\Rightarrow \) a² = \(k\over4\)
A path difference of \(\lambda\over3\), corresponds to a phase diference of \(2\pi\over3\)
\(\therefore\)Intensity = 4a² cos2 \(\phi\)/2
= 4 x a² x cos² \({2\pi/3}\over2\)
= 4 x \(k\over4\) x \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)= \(k\over4\)

(ii) For \(\lambda _{ 1 }=590nm\)  
 Location of 1st maxima,\(\lambda _{ 1 }=\left( 2n+1 \right) \frac { D\lambda _{ 1 } }{ 2a } \) 
If  \(n=1 \ \Rightarrow \lambda _{ 1 }=\frac { 3D\lambda _{ 1 } }{ 2a } \)
For \(\lambda _{ 2 }=596nm\)
Location of 2nd maxima,\(\lambda _{ 2 }=\left( 2n+1 \right) \frac { D\lambda _{ 1 } }{ 2a } \)
If  \(\lambda _{ 2 }=\left( 2n+1 \right) \frac { D\lambda _{ 1 } }{ 2a } \)
Path difference \(\lambda _{ 2 }-\lambda _{ 1 }=\frac { 3D }{ 2a } \left( \lambda _{ 2 }-\lambda _{ 1 } \right) \)
\(=\frac { 3\times 1.5 }{ 2\times 2\times { 10 }^{ -6 } } \left( 596-590 \right) \times { 10 }^{ -9 }\)
\(=6.75\times { 10 }^{ -3 }m\)

(i) \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ R } \)      ............(i)
(ii) Now, the image I^' acts as a virtual object for the second surface that will form a real at I. As, refraction takes place from denser to rarer medium,

\(\therefore \quad \frac { { -n }_{ 2 } }{ v } +\frac { { n }_{ 1 } }{ { v }^{ ' } } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }^{ ' } } \)   ............(ii)
On adding Eqs. (i) and (ii), we get
\(\frac { 1 }{ f } =\left( { n }_{ 21 }-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ { R }^{ ' } } \right) \ \left[ \therefore { n }_{ 21 }=\frac { { n }_{ 2 } }{ { n }_{ 1 } } ,\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \right] \)

(d): When a light wave travel from a rarer to a denser medium it loses speed, but energy carried by the wave does not depend on its speed. Instead, it depends on the amplitude of wave. The frequency also remain constant.

(b): For case when dark fringe is observed opposite to one of the slit
\(\text { here } S_{1} P=D \text { and } S_{2} P=\sqrt{D^{2}+d^{2}}=D\left[1+\frac{d^{2}}{2 D^{2}}\right]\)
\(\text { Path difference }=S_{2} P-S_{1} P=\frac{d^{2}}{2 D}=\frac{\lambda}{2} \text { or, } \lambda=\frac{d^{2}}{D}\)