(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

(a) Refractive index of glass, μ = 1.5
Speed of light, c = 3 x 10⁸ m/s
Speed of light in glass is given by the relation
\(v=\frac{c}{\mu}\)
\(=\frac{3 \times 10^{8}}{1.5}=2 \times 10^{8} \mathrm{~m} / \mathrm{s}\)
Hence, the speed of light in glass is 2 x 10⁸ m/s.
(b) The speed of light in glass is not independent of the colour of light
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

\(Here, \ \lambda =589 \ nm,\ c=3\times { 10 }^{ 8 }m/s, \ \mu =1.33\)
(a) For reflected light
\(wavelength,\ \lambda =589\quad nm=589\times { 10 }^{ -9 }m,\quad v=\frac { c }{ \lambda } =\frac { 3\times { 10 }^{ 8 } }{ 589\times { 10 }^{ -9 } } =5.09\times { 10 }^{ 14 }hertz\)
\(speed,\ v=c=3\times { 10 }^{ 8 }m/s\)
(b) For refracted light \( \lambda '=\frac { \lambda }{ \mu } =\frac { 589\times { 10 }^{ -9 } }{ 1.33 } =4.42\times { 10 }^{ -7 }m\)
As frequency remains unaffected on entering another medium,
\(\\ therefore,\quad v'=v=5.09\times { 10 }^{ 14 }hertz\)
\(speed, \ v'=\frac { c }{ \mu } =\frac { 3\times { 10 }^{ 8 } }{ 1.33 } =2.25\times { 10 }^{ 8 }m/s\)

Interference of Light. The phenomenon of redistribution of energy in a medium due to superimposition of waves from two coherent source of light is called Interference of Light.

Constructive Interference.  At points, where the crest of one wave falls the crest of the other or a through of one falls on the through of the other, the amplitude of the resulting wave becomes maximum. Hence the energy or the intensity of light at such points becomes maximum. This is called Constructive Interference.

Destructive Interference. At some other points where the through of one falls on the crest of the other or crest of one falls on the through of the other, the amplitude of the resulting waves becomes minimum. Hence the energy or intensity becomes minimum. This is called Destructive Interference.

Law of conservation of energy is obeyed. It should be clearly understood that in interference of light no light energy is destroyed. The loss of energy at the points of destructive interference appears as the increase of energy at the points of constructive interference.

(a) (i) Ray of light should travel from denser to rarer medium.
(ii) Angle of incidence should be more than the critcical angle 
(b) \(\mu =\frac { 1 }{ { sini }_{ c }\quad } \) where i_c is the critical angle

3 cm

For a given frequency, intensity of light in the photon is determined by the number of photons crossing a unit area per unit time.

Yes, because objective with larger diameter will collect more light and even the distant objects can be seen.

(a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators, which take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus the frequency of scattered light equals the frequency of incident light.
(b) No, Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation.
(c) For a given frequency, intensity of light in the photon crossing an unit area per uni time.

(i) Given, the focal length of convex lens, f₁ = 30 cm
The focal length of concave lens, f₂ = -20 cm, d = 8 cm
Use the formula,
 \(\frac { 1 }{ f } =\frac { 1 }{ { f }_{ 1 } } +\frac { 1 }{ { f }_{ 2 } } -\frac { d }{ { f }_{ 1 }{ f }_{ 2 } }\)
\(\frac { 1 }{ f } =\frac { 1 }{ 30 } -\frac { 1 }{ 20 } -\frac { 8 }{ 30\times (-20) } =\frac { 20-30+8 }{ 30\times 20 }\)
\( \frac { 1 }{ f } =-\frac { 2 }{ 600 } \)
\( f=-300 \ cm\)
(a) Let us take that the incident beam falls on convex lens and assume that concave lens is absent.
 \(\therefore \ { u }_{ 1 }=\infty ,{ f }_{ 1 }=30 \ cm\)
\(Using \ lens \ formula,\frac { 1 }{ { f }_{ 1 } } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } \)
\(\\ d \frac { 1 }{ 30 } =\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ \infty } \)
Position of the image formed by convex lens,
 v₁= 30 cm
Now, this image acts as an object for concave lens.
Now, distance of object
u₂ = + (30-8) = 22 cm
f₂ = -20 cm,
 v₂ = ?
   
Using lens formula,
 \(\frac { 1 }{ { f }_{ 2 } } =\frac { 1 }{ { v }_{ 2 } } -\frac { 1 }{ { u }_{ 2 } } \)
\(\frac { 1 }{ { v }_{ 2 } } =-\frac { 1 }{ 20 } +\frac { 1 }{ 22 } =-\frac { 1 }{ 220 }\)
Distance of final image, v₂ = -220 cm
Thus, the parallel neam would appear to diverge from a point 220-4 = 216 cm from the centre of two lens system.
(b) Let us take the parallel beam first falls on concave lens.     
  
\({ u }_{ 1 }=-\infty ,{ f }_{ 1 }=-20cm,{ v }_{ 1 }=?\)
\( \frac { 1 }{ { v }_{ 1 } } +\frac { 1 }{ \infty } =\frac { 1 }{ -20 } \)
\( { v }_{ 1 }=-20 \ cm\)
This acts as an object for convex lens.
\({ u }_{ 2 }=-(20+8)=-28cm,\)
\( { f }_{ 2 }=30cm,\ { v }_{ 2 }=?\)
\(\frac { 1 }{ { v }_{ 2 } } +\frac { 1 }{ 28 } =\frac { 1 }{ 30 } \)
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ 30 } -\frac { 1 }{ 28 } =\frac { 14-15 }{ 420 } =-\frac { 1 }{ 420 }\)
\({ v }_{ 2 }=-420cm\)
The parallel beam appears to diverge from a point 420-4 = 416 cm on the left of the centre of the two lens system. From the above two cases it concludes that the answer depends on which side of the lens system, the parallel beam is incident. So, the notion of effective focal length does not seem to be useful here.
(ii) Give, size of object, O = 1.5 cm
Distance from the convex lens, u₁= -40 cm
For the convex lens, \(\frac { 1 }{ { v }_{ 1 } } -\frac { 1 }{ { u }_{ 1 } } =\frac { 1 }{ { f }_{ 1 } }\)
  
So, \(\frac { 1 }{ { v }_{ 1 } } =\frac { 1 }{ 30 } +\frac { 1 }{ (-40) } =\frac { 4-3 }{ 120 } =\frac { 1 }{ 120 } \)
v₁ = 120 cm
Magnification produced by convex lens,
\({ m }_{ 1 }=\frac { { v }_{ 1 } }{ { u }_{ 1 } } =\frac { 120 }{ -40 } =-3\)
Now, this image acts as an object for concave lens, 
 For concave lens, u₂ = 120 - 8 = 112 cm,
f₂ = -20 cm
So,
\(\frac { 1 }{ { v }_{ 2 } } =\frac { 1 }{ -20 } +\frac { 1 }{ 112 } =\frac { -112+20 }{ 112\times 20 } =\frac { -92 }{ 112\times 20 }\)
\({ v }_{ 2 }=\frac { -112\times 20 }{ 92 } cm\)
Magnification produced by concave lens,
\({ m }_{ 2 }=\frac { { v }_{ 2 } }{ { u }_{ 2 } } =\frac { (-112\times 20) }{ 92\times 112 } =-\frac { 20 }{ 92 }\)
Maginification produced by combination,
\(m={ m }_{ 1 }\times { m }_{ 2 }=(-3)\times (-\frac { 20 }{ 90 } )=\frac { 60 }{ 92 } \)
m = 0.652
Size of image = m x Size of object
 = 0.652 x 1.5 = 0.98 cm
Thus. the magnification produced by the two lens system is 0.652 and size of image is 0.98 cm.

radii of curvatures R₁ = 9 cm , R₂ = -15 cm
  \(\frac { 1 }{ f } =(\mu -1)\frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } }\)
\( \\ \frac { 1 }{ f } =(1.5-1)\frac { 1 }{ 9 } +\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ f } =0.5(\frac { 45 }{ 4 } )=11.25\quad cm\)
\(\\ \frac { 1 }{ f } =\frac { 1 }{ 11.25\times 10^{ -2 } } \)
\(p=\frac { 1000 }{ 1125 } =8.88D\)

Angular magnification is maximum, when final image is formed at the least of distinct vision.
v = -25cm, \(f={100\over P}={100\over10}=10cm, u=?\)
From  \({1\over v}-{1\over u}={1\over f}\)
\({1\over u}={1\over v}-{1\over f}={1\over -25}-{1\over 10}={-7\over50}\)
\(u=-{50\over7}\)
= -7.1cm

Let I₀ be the intensity of polarised light after passing through the first polariser P₁. Then the intensity of light after passing through second polariser P₂ will be
I = I₀ cos ² \(\theta\) 
where \(\theta\) is the angle between pass axes of P₁ and P₂. Since P₁ and P₃ are crossed the angle between the pass axes of P₂ and P₃ will be ( \(\pi\) / 2 –\(\theta\) ). Hence the intensity of light emerging from P₃ will be
\(I=I_{0} \cos ^{2} \theta \cos ^{2}\left(\frac{\pi}{2}-\theta\right)\)
\(=I_{0} \cos ^{2} \theta \sin ^{2} \theta=\left(I_{0} / 4\right) \sin ^{2} 2 \theta\)
Therefore, the transmitted intensity will be maximum when \(\theta\) = \(\pi\) / 4.

(i) (c) :  Given   \(I_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \text { and } I_{2}=25 \mathrm{~W} / \mathrm{m}^{2}\)
\(\frac{I_{1}}{I_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{10}{25} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{3.16}{5} \text { or } a_{1}=\frac{3.16}{5} a_{2}=0.6324 a_{2}\)
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{\left[0.6324 a_{2}+a_{2}\right]^{2}}{\left[0.6324 a_{2}-a_{2}\right]^{2}}=19.724\)
(ii) (b): In an excessively thin film, the thickness of the film is negligible. Thus the path difference between the reflected rays becomes \(\lambda\)/2 which produces a minima.
(iii) (a): Since, \(\beta=\frac{\lambda D}{d} \text { for } d=2 d\)
\(\beta^{\prime}=\frac{\lambda D^{\prime}}{2 d}=\beta(\text { Gives })\)
\(\therefore \quad D_{1}=2 D\)
(iv) (b): The condition for possible interference maxima on the screen is, dsin \(\theta\) = n\(\lambda\)
where d is slit separation and Ais the wavelength.
\(\text { As } d=2 \lambda \text { (given) } \quad \therefore 2 \lambda \sin \theta=n \lambda \text { or } 2 \sin \theta=n\)
For number of interference maxima to be maximum,sin\(\theta\) = 1 :. n = 2
The interference maxima will be formed when n = 0, ± 1, ± 2
Hence the maximum number of possible maxima is 5.
(v) (d): \(y_{1}=a \sin \left(\omega t+\frac{\pi}{3}\right) \text { and } y_{2}=a \sin \omega t\)
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}, \text { where } \phi=\frac{\pi}{3}\)
\(=\sqrt{a^{2}+a^{2}+2 a a \cos \frac{\pi}{3}}=\sqrt{3} a\)

(c) A is true but R is false.

(c) A is true but R is false.

(c): Fringe width \(\beta\) = \(\lambda\)D/d shall remain the same as the waves travel in air only, after passing through the thin transparent sheet. Due to introduction of thin sheet, only path difference of the wave is changed due to which there is shift of position of fringes only, Which is given As \(\Delta x=\frac{D(\mu-1) t}{d}\)Where \(\mu\) is refractive index of thin sheet and t is its thickness.