When a plane wavefront is incident on a single slit, all the point sources of light constituting the wavefronts are in same phase. The wavelets coming out from the wavefront might meet over the screen with some path difference, i.e. a phase difference is introduced between them.

The brightness at a point on the screen depends on the phase difference between the wavelets meeting at the point. We imagine that the slit is divided into smaller parts and the wavelets coming out from these portions meet and superpose on the screen with proper phase difference.
The wavelets from different parts of the wavefront, incident on the slit, meet with zero phase difference to constitute a central maximum. In case of secondary maxima, there are some wavelets meeting the screen out of phase, thus, reducing intensity of secondary maxima.

(i) Coherent sources produce light of constant phase difference and hence, permanent fringr pattern is produced due to interfernce of light.
(iii) Intensity of light at a point on the screen is given by
\({ I }_{ R }={ I }_{ 1 }+{ I }_{ 2 }+2\sqrt { { I }_{ 1 }{ I }_{ 2 }cos\phi } \)
For the path difference \(\lambda \) , phase difference is 2\(\pi \).
As, sources are coherent and taken out of the same source in Young's double slit experiment,
 \({ I }_{ 1 }={ I }_{ 2 }=I\)
\(\\ { I }_{ R }=2I+2I \ cos2\pi \) 
\(\\ { I }_{ R }=4I\)
\(\\ 4I=K\quad unit\quad \quad \quad \quad \quad .........(1)\)
For path difference, \(\frac { \lambda }{ 3 } \) correspomding to phase difference of \(\frac { 2\pi }{ 3 } \)
\({ I }_{ R }=2I+2I\quad cos\frac { 2\pi }{ 3 } =2I-I=\quad I\quad \quad \quad \quad \quad \quad .......(ii)\)
From Eqs. (i) and (ii), we conclude
\({ I }_{ R }=\frac { K }{ 4 } unit\)​​​​

Here, the point P on the right of the lens acts as a virtual object
u = 12 cm, v = ?
(a) f = 20 cm
\(As \ \frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }\)
\( \frac { 1 }{ v } -\frac { 1 }{ 12 } =\frac { 1 }{ 20 } \)
\( \frac { 1 }{ v } =\frac { 1 }{ 20 } +\frac { 1 }{ 12 } =\frac { 3+5 }{ 60 } =\frac { 8 }{ 60 } \)
\( v=60/8=7.5cm\)
The image is at 7.5cm to the right of the lens where the beam converges.
(b) f = -16 cm, u = 12 cm
\(As\quad \frac { 1 }{ v } =\frac { 1 }{ f } +\frac { 1 }{ u }\)
\( -\frac { 1 }{ 16 } +\frac { 1 }{ 12 } =\frac { -3+4 }{ 48 } =\frac { 1 }{ 48 }\)
\( v=48 \ cm\)
Hence, the image is at 48 cm. to the right of the lens, where the beam would converge.

\(First\ case,\)
\(angle \ of \ incidence \ i={ 60 }^{ \circ }\)
\(angle \ of \ refraction \ r={ 35 }^{ \circ }\)
\({ \alpha }_{ { \mu }_{ g } }= \ \frac { 1 }{ { g }_{ { \mu }_{ \alpha } } } =\frac { sin\ i }{ sin \ r } \)
\(Second \ case,\)
\( { \alpha }_{ { \mu }_{ \omega } }=\frac { sin\ { 60 }^{ \circ } }{ sin\ { 47 }^{ \circ } } =1.18\)
\( { \omega }_{ { \mu }_{ g } }= { \alpha }_{ { \mu }_{ g } }\times { \omega }_{ { \mu }_{ \alpha } }\)
\(=\frac { { \alpha }_{ { \mu }_{ g } } }{ { \alpha }_{ { \mu }_{ \omega } } } =\frac { 1.51 }{ 1.18 } =1.28\)
\(Third \ case,\)
\( angle \ of \ incidence \ i={ 45 }^{ \circ }\)
\( angle \ of \ refraction \ r=?\)
\( { \omega }_{ { \mu }_{ g } }=\frac { sin\ i }{ sin\ r } \)
\(\frac { sin \ i }{ sin\ r } =1.28\)
\(sin \ r=\frac { sin \ { 45 }^{ \circ } }{ 1.28 } =0.5525\)
\( sin \ r= \ sin{ 33 }^{ \circ }54'\)
\(r= \ { 33 }^{ \circ }54'\)

Lens maker's formula 

(i) light waves must have same frequency/ wavelengths
(ii) they should have zero or constant phase diff.

Yes, provided the medium is homogeneous.

Change in the velocity of light on change in the medium.

(a)

(b) \({ y }_{ n }=\frac { n\lambda D }{ d } \)
\(d=\frac { n\lambda D }{ { y }_{ n } }\)
\( =\frac { 1\times 500\times { 10 }^{ -9 }\times 1 }{ 2.5\times { 10 }^{ -3 } } m\)
\( =2\times { 10 }^{ -4 }m(=0.2mm)\)

(a) Both A and R are true and R is the correct explanation of A.

(a): The central spot of Newton's rings is dark when the medium between plano convex lens and plane glass is rarer than the medium of lens and glass. The central spot is dark because the phase change of \(\pi\) is introduced between the rays reflected from surfaces of denser to rarer and rarer to denser media.

(i) n the double-slit experiment, the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole and the double-slit interference pattern as shown in the given figure. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a, which is the ratio of the distance between the two slits to the width of a slit. In the limit of a becoming very small, the diffraction pattern will become flat and we will observe the two slit interference pattern.

(ii) (a) \(\lambda\)Given  = 400 nm
= 4 x 10^{- 7} m,
d = 20 \(\mu\)m = 2 x 10^{- 5} m,
a = 4\(\mu\)m = 4 x 10^-6 m
If n = no. of bright fringes within central peak of diffraction envelope
Then \(\frac{2 \lambda \mathrm{D}}{a}=n \lambda \frac{\mathrm{D}}{d}\)
\(\Rightarrow \ \frac{2}{a}=\frac{n}{d}\)
\(\Rightarrow \quad n=\frac{2 d}{a}=\frac{2 \times 2 \times 10^{-5}}{4 \times 10^{-6}}=10\)
(b) \(\frac{\lambda \mathrm{D}}{a}=n \lambda \frac{\mathrm{D}}{d}\)
\(\Rightarrow \ \frac{1}{a}=\frac{n}{d}\)
\(\Rightarrow \ n=\frac{d}{a}=\frac{2 \times 10^{-5}}{4 \times 10^{-6}}=5\)​​​​