λ = 500 nm = 500 x 10^-9 m
x = 3 mm = 3 x 10^-3 m
a = 1 mm = 1 x 10^-3 m
\(d=\frac{xa}{1.22 \lambda}\)
\(d=\frac{3 \times1\times10^{-6}}{1.22 \times500\times10^{-9}}\)
\(=\frac{3 \times1\times10^{-6}}{6.10 \times 10^{-7}}\)
\(d=\frac{30}{6.1}=5 m\)

Convex Mirror is diverging in nature and for all positions of objects, convex mirror forms virtual and erect image.

\(\\ \mu =\cfrac { c }{ v } =\cfrac { sini }{ sinr } \) 
It follows that: v of sin r Since r is minimum in medium R, therefore, sin r and hence the velocity of light is minimum in medium R.

(i) Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging.
(ii) In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

The angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion.

Yes, there is a difference between coloured light obtained from prism and soap bubble Dispenion takes place in prism. Interference takes place in soap bubbles.

n = 1.65
Brewster’s law, tan i_p= n
tani_p = 1.65, i_p = tan^-11.65, i_p = 58.5^o
The inclination with the horizontal surface is (90^o - 58.8^o) = 31.2^o

The refractive index of vacuum, n₁ = 1
The wavelength in vacuum, λ₁ = 5893 Å.
The speed in vacuum, c = v₁ = 3 x 10⁸ m s^–1
The refractive index of water, n₂ = 1.33
The wavelength of light in water, λ₂
The speed of light in water, v₂
(a) The equation relating the wavelength and refractive index is,
\(\cfrac { { \lambda }_{ 1 } }{ \lambda _{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Rewriting, \({ \lambda }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { \lambda }_{ 1 }\)
Substituting the values,
 \({ \lambda }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 5893\overset { o }{ A } =4431\overset { o }{ A } \) 
\({ \lambda }_{ 2 }=4431\overset { o }{ A } \) 
(b) The equation relating the speed and refractive index is,
\(\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Rewriting, \({ v }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { v }_{ 1 }\) 
Substituting the values,
\({ v }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 3\times { 10 }^{ 8 }=2.256\times { 10 }^{ 8 }\) 
v₂ = 2.256 x 10⁸ ms^-1
(c) Frequency of light in vacuum is,
\({ v }_{ 1 }=\cfrac { c }{ { \lambda }_{ 1 } } \) 
Substituting the values,
\({ v }_{ 1 }=\cfrac { 3\times { 10 }^{ 8 } }{ 5893\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
Frequency of light in water is, \({ v }_{ 2 }=\cfrac { v }{ { \lambda }_{ 2 } } \)
Substituting the values,
 \(\\ { v }_{ 2 }=\cfrac { 2.256\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 4431\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
The results show that the frequency remains same in all media.

Since, the prism is equilateral, A = 60^o;
Given, i₁ = 30^o;i₂ = 75^o
Equation for angle of deviation, d = i₁ + i₂ – A
Substituting the values, d = 30°+ 75°– 60°= 45°
The angle of deviation produced d = 45^o

Let us assume a person of height h is standing in front of a vertical plane mirror. The person could see his/her head when light from the head falls on the mirror and gets reflected to the eyes. Same way, light from the feet falls on the mirror and gets reflected to the eyes.

If the distance between his head H and eye E is h₁ and distance between his feet F and eye E is h₂. The person’s total height is, h Here it is, h = h₁ + h₂ 
By the law of reflection, the angle of incidence and angle of reflection are the same in the two extreme reflections. The normals are now the bisectors of angles between incident and reflected rays in the two points. By geometry, the height of the mirror needed is only half of the height of the person. \( \frac { { h }_{ 1 }+{ h }_{ 2 } }{ 2 } =\frac { h }{ 2 } \).
Does the height depend on the distance between the person and the mirror?

The angle between the incident ray and deviated ray of light ray is called Angle of deviation due to reflection. d = 180 - 2i

When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell's window

Mirage:
Mirage is an optical illusion caused by atmospheric conditions especially the appearance of sheet of water in a desert caused by total internal reflection (or) refraction of light from the sky by heated air.
Looming:
Looming is an optical illusion caused by bending of light which appear an object floating high above its actual position specially in polar region.

Stars appear twinkling because of the movement of the atmospheric layer with varying refractive indices due to refraction.

The angle between the incident and deviated light is called Angle of deviation due to refraction. When light travels from
(i) rarer to denser medium, d = i - r
(ii) denser to rarer medium, d = r - i

\(sin{ i }_{ c }=\cfrac { 1 }{ \mu } =\cfrac { 1 }{ 1.5 } =0.667\\ \) 
or i_c= 41.8^o
Deviation = 90^o - i = 90^o - 41.8^o = 48.2^o
This si the maximum attainable deviation in refraction. So, the given data favours total internal reflection.
In reflection, deviation = 180^o - 2i
or 90^o = 180^o - 2i or 2i = 90^o or i = 45^o
or i = 45^o

For minimum deviation position
\({ \mu }_{ red }=\cfrac { sin\left( \frac { A+{ \delta }_{ red } }{ 2 } \right) }{ sin\frac { A }{ 2 } } \) 
or \(sin\left( \cfrac { A+{ { \delta }_{ red } } }{ 2 } \right) ={ n }_{ red }\) 
\(sin\cfrac { 2 }{ A } =1.622\times 0.5=0.811\) 
\(\therefore \cfrac { 60+{ \lambda }_{ red } }{ 2 } ={ 54 }^{ o }12'\) 
\({ \delta }_{ red }={ 108 }^{ o }24'-{ 60 }^{ o }={ 48 }^{ o }24'\)
Similarly, \(sin\left( \cfrac { A+{ \delta }_{ viloet } }{ 2 } \right) =1.663\times 0.5'\) 
= 0.8315 or \(\cfrac { { 60 }^{ o }+{ \delta }_{ viloet } }{ 2 } =56^{ o }15'\) 
\({ \delta }_{ viloet }=112^{ o }30'-{ 60 }^{ ' }={ 52 }^{ o }30'\) 
\(\therefore { \delta }_{ violet }-{ \delta }_{ red }=\left( 52-30' \right) -\left( 48^{ o }24' \right) ={ 4 }^{ o }6'\) 
It is not advisable to use the formula,
\({ \delta }_{ v }-{ \delta }_{ r }=\left( { \mu }_{ v }-{ \mu }_{ r } \right) \) in the above solution.

P = P₁ + P₂ = (15 - 5) D = 10D
\(F=\cfrac { 1 }{ P } =\cfrac { 1 }{ 10 } m=10cm\) 
(b) \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ F } or\cfrac { 1 }{ F } +\cfrac { 1 }{ u } \) 
or \(\cfrac { 1 }{ v } =\cfrac { 1 }{ 10 } +\cfrac { 1 }{ -30 } \) or v = 15cm
Again,\(\cfrac { I }{ O } =\cfrac { v }{ u } =\cfrac { 15 }{ -30 } =-\cfrac { 1 }{ 2 } \) 
or \(I=-\cfrac { 1 }{ 2 } \times 3cm=-1.5cm\) 
The negative sign indicates that the image is real and inverted.

For face AB, ∠i = 0°,∠r = ·0° the ray will pass through AB undeflected 
Now, at face AC

Formula.
Here,\({ \quad i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ \mu } \right) \) 
= \({ sin }^{ -1 }\left( \cfrac { 2 }{ 3 } \right) ={ sin }^{ -1 }\left( 0.66 \right) \) 
∠i on face AC is 30^o which is less than ∠i_e.
Hence, the ray get refracted.
And, applying Snell's law at face AC
\({ sin30 }^{ o }\times \cfrac { 3 }{ 2 } =sinr\times 1\) 
\(\Rightarrow sinr=\cfrac { 1 }{ 2 } \times \cfrac { 3 }{ 2 } \) 
\(\Rightarrow r={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) ={ sin }^{ -1 }(0.75)\)
And, dearly r > i, as ray passes from denser to rarer medium.

(i) At a minimum deviation \(\mu =\cfrac { sin\left( \frac { A+{ \delta }_{ m } }{ 2 } \right) }{ sin\left( \cfrac { A }{ 2 } \right) } \) 
 Given \({ \delta }_{ m }=A\)
\(\mu =\cfrac { sinA }{ sin\cfrac { A }{ 2 } } =\cfrac { 2sin\cfrac { A }{ 2 } cos\cfrac { A }{ 2 } }{ sin\cfrac { A }{ 2 } } \) 
= \(2cos\cfrac { A }{ 2 } \) 
\(\therefore cos\cfrac { A }{ 2 } =\cfrac { \sqrt { 3 } }{ 2 } \cfrac { A }{ 2 } =30\) A = 6^o 
(iii) \(\mu =\sqrt { 3 } \cfrac { 1 }{ { sini }_{ c } } \Rightarrow { sini }_{ c }=\cfrac { 1 }{ \sqrt { 3 } } \) 
ஃ Angle of incidence > i_c
Total internal reflection takes place.

Dispersion: It is splitting of white light into its constituent colours.
(i) Consider a beam of white light passes through a prism; it gets dispersed into its constituent colours as shown in Figure.
(ii) Let \(\delta_{v}, \delta_{R} \) are the angles of deviation for violet and red light. Let n_V and n_R are the refractive indices for the violet and red light respectively
(iii) The refractive index of the material of a prism is given by the equation
\(\mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{D}}{2}\right)}{\sin (\mathrm{A} / 2)}\)
(iv) Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10^o, the prism is said to be a small angle prism.
(v) When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a smitt angle prism and the angle of deviation then the prism formula becomes.

\(n=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin (A / 2)}\)
For small angles of  \(A\ and \ \delta \)
\(\sin \frac{A+\delta}{A} \approx \frac{A+\delta}{A} \)
\(\sin \frac{A}{2} \approx \frac{A}{2} \)
\(n=\frac{(A+\delta / 2)}{(A / 2)}=\frac{A+\delta}{A}=1+\frac{\delta}{A} \)
Further simplifying, 
\(\frac{\delta}{A} =n-1 \)
\(\delta =(n-1) A \)    .....(1)
(vi) When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours
For Violet colour, \( \delta_{\mathrm{v}}=\left(\mathrm{n}_{\mathrm{v}}-1\right) \mathrm{A} \)   ...(2)
For Red colour, \(\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{R}}-1\right) \mathrm{A} \)   ....(3)
(vii) As, angle of deviation for violet colour \(\delta_{v}\) is greater the angle of deviation for red colour \(\delta_{\mathrm{R}}\) the refractive index for violet colour n_v is greater than the refractive index for red colour n_R Subtracting \(\delta_{v}\)  from \(\delta_{\mathrm{R}}\) we get
\(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}\right) \mathrm{A}\) ....(4)
(viii) The term \(\left(\delta_{v}-\delta_{R}\right)\)  is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. If we take \(\delta\) is the angle of deviation for any middly ray (green or yellow) and the corresponding refractive index. Then,
\(\delta=(n-1) A\)    .....(5)
Dispersive power (ω):
It is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power
\(\omega=\frac{\text { Angular dispersion }}{\text { Mean deviation }}=\frac{\delta_{v}-\delta_{R}}{\delta}\)  ....(6)
Substituting \(\left(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}\right) \text { and }(\delta) \)
\(\omega=\frac{\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}}{(\mathrm{n}-1)} \)  .......(7)
(ix) Dispersive power is a dimensionless quality It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

Angle of deviation Produced by Prism:
(i) Let light ray PQ is incident on one of the refracting faces of the prism.
(ii) The angles of incidence and refraction at the first face AB are i₁ and r_l. The path of the light inside the prism is QR.
(iii) The angle of incidence and refraction at the second face AC is r₂ and i₂ respectively.
(iv) RS is the ray emerging from the second face. Angle i₂ is also caned angle of emergence.
(v) The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
(vi) The two normals drawn at the point of incidence Q and emergence R meet at point N. They meet at point N.
(vii) The extended incident ray and the emergent ray meet at a point M.

The angle of deviation d₁ at the surface AB is,
ㄥRQM = d = i₁ - r₁ ...(1)       
The angle of deviation d₂ at the surface AC is
ㄥQRM = d₂ = i₂ - r₂          .......(2)
Total angle of deviation d produced is, 
d = d₁ + d₂             .....(3)
Substituting for d₁ and d₂ in equation (3)
d = (i₁ - r₁) + (i₂ - r₂)
After rearranging,
d = (i₁ - r₁) + (i₂ - r₂)                    ........(4)
 In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
\(\angle A+\angle QNR={ 180 }^{ 0 }\)         .........(5)
From the triangle ΔQNR 
\({ r }_{ 1 }+{ r }_{ 2 }+\angle QNR={ 180 }^{ o }\)                ......(6)
Comparing these two equations (5) and (6) we get,
r₁ + r₂ = A              .......(7)
Substituting this in equation (4) for angle of deviation,
d = i₁+ i₂ - A                .............(8)
(viii) Thus, the angle of deviation depends on the angle of incidence i₁, angle of emergence i₂ and the angle for the prism A. 
(ix) For a given angle of incidence the angle of emergence is decided by the refractive index of the material of the prism. Hence the angle of deviation depends on these following factors.
(i) the angle of incidence
(ii) the angle of the prism.
(iii) the refractive index of the material of the prism (which decides the angle of emergence).
Refractive index of the material of the prism:

At minimum deviation, i₁ = i₂ = i and r₁ = r₂ = r
Now, the equation (8) becomes,
D - i₁ + i₂ - A = 2_i - A (or) \(i=\cfrac { \left( A+D \right) }{ 2 } \) 
The equation (7) becomes
r₁ + r₂ = A ⇒ 2r = A (or) \(r=\cfrac { A }{ 2 } \)
Substituting i and r in Snell's law
\(n=\cfrac { sini }{ sinr } \) 
\(n=\cfrac{\cfrac{sin(A+D)}{2}}{sin(A/2)}\)

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.

(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (1) forms image at 1'.
(iii) Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given from Equation,
 \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ v} =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R } } \)
For the refracting surface (1), the light goes from n₁ to n₂
\(\cfrac { { n }_{ 2 } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1 } } \)   .....(1)
For the refracting surface (2), the light goes from n₂ to n₁
\(\frac{n_{1}}{v}+\frac{n_{2}}{v^{\prime}}=\frac{\left(n_{1}-n_{2}\right)}{R_{2}}\)    ......(2)
For surface (2) I' acts as virtual object.
Adding the above two equations (1) and (2)
\(\cfrac { { n }_{ 1 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
on further simplifying and rearranging,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ n_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ....(3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \(\infty\), v = f. Then the equation becomes.
\(\cfrac { 1 }{ f } -\cfrac { 1 }{ \infty } =\left( \cfrac { { n }_{ 2 } }{ { { n }_{ 1 } } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ......(4)
If the lens is kept in air, then we can take n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\\ \cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ..(5)
The above equation is called the lens maker's formula.
Significance:
It tells the lens manufacturers what curvature is needed to make a lens of desired focal length with a material of particular refractive index to make a lens of desired focal length. This formula holds good also for a concave lens.

Apparatus:
The apparatus used by Fizeau for determining speed of light in air is shown in Figure 

(i) The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light.
(ii) The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.
(iii) The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel.
(iv) The toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer, through the partially silvered glass plate.
Working:
(i) The angular speed of rotation of the toothed wheel was increased from zero to a value w until light passing through one cut would completely be blocked by the adjacent tooth.
(ii) This is ensured by the disappearance of light while looking through the partially silvered glass plate.
Expression for speed of light:
The speed of light in air 'v' is equal to the ratio of the distance 2d (the distance light travelled from the toothed wheel to the mirror and back) to the time taken 't'.
\(v=\cfrac { 2d }{ t } \)   ...(i)
(v) The distance d is a known value from the arrangement. The time taken 't' for the light to travel the distance (2d) is calculated from the angular speed 'ω' of the toothed wheel.
(vi) The angular speed 'ω' of the toothed wheel when the light disappeared for the first time is,
\(\omega =\cfrac { \theta }{ t } \)  ...(2)
(vii) Here, θ is the angle between the tooth and the next slot which is turned within that time t.
\(\theta =\cfrac { total \ angle\ of \ the\ circle \ in \ radian }{ number\ of\ teeth+number\ of\ cuts } \)
\(\theta =\cfrac { 2\pi }{ 2N } =\cfrac { \pi }{ N } \) 
Substituting for θ in the equation for ω,
\(\omega =\cfrac { \pi /N }{ t } =\cfrac { \pi }{ Nt } \) 
Rewriting the above equation for t,
\(t=\cfrac { \pi }{ N\omega } \)    ....(3) 
Substituting 't' in equation (1)
\(V=\cfrac { 2d }{ \pi /N\omega } \)  
After rearranging,
\(v=\cfrac { 2dN\omega }{ \pi } \)
From this, the speed of light in air was determined as, v = 2.997 x 10⁸ m/s.

Yes, It is possible for two lens to produce zero power.
Explanation:
when the two lens (one is concave & another one is convex lens) are combined together, the focal length of the combination of the two lenses is F.
Then
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)
\(\text { if } f_{1}=f_{2}=f \)
\(f_{1}=f(\text { convex }) \)
\(f_{2}=-f(\text { concave) } \)
\(P=\frac{1}{F}=\frac{1}{f}-\frac{1}{f}=0 . \ P \rightarrow \text { power } \)

λ = 589 nm = 589 x 10⁻⁹ m
For the film to have strong reflection, the reflected waves should interfere constructively. The least optical path difference introduced by the film should be λ/2. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. Thus, for strong reflection, 2μd = λ/2 [As given in equation 6.145. with n = 1]
Rewriting, \(d=\frac{\lambda}{4 \mu}\) 
Substituting, \(d=\frac{589 \times 10^{9}}{4 \times 1.25}=117.8 \times 10^{-9}\) 
d = 117.8 x 10^-9 = 117.8 nm
For the film to be anti-reflecting, the reflected rays should interfere destructively. The least optical path difference introduced by the film should be λ. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. For strong reflection, 2μd = λ [As given in equation 6.146. with n = 1]
Rewriting, \(d=\cfrac { \lambda }{ 2\mu } \)
Substituting, \(d=\cfrac { 589\times { 10 }^{ 9 } }{ 2\times 1.25 } =235.6\times { 10 }^{ -9 }\) 
d = 235.6 x 10^-9 = 235.6 nm

Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude,
\(A=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+2{ a }_{ 1 }{ a }_{ 2 }cos\varphi } \) 
Resultant amplitude is maximum when,
\(\phi =0,cos0=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \) 
 \(\\ { A }_{ max }=\sqrt { \left( { a }_{ 1 }+{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5+3 \right) ^{ 2 } } =\sqrt { \left( 8 \right) ^{ 2 } } \) 
= 8 units
Resultant amplitude is minimum when
\(\phi =\pi,cos\pi=1,{ A }_{ max }=\sqrt { { a }_{ 1 }^{ 2 }+{ a }_{ 2 }^{ 2 }+{ 2a }_{ 1 }{ a }_{ 2 } } \)
\({ A }_{ max }=\sqrt { \left( { a }_{ 1 }-{ a }_{ 2 } \right) ^{ 2 } } =\sqrt { \left( 5-3 \right) ^{ 2 } } =\sqrt { \left( 2 \right) ^{ 2 } } \)
= 2units
\(I\infty { A }^{ 2 }\) 
\(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( { A }_{ max } \right) ^{ 2 } }{ \left( { { A }_{ min } } \right) ^{ 2 } } \)
Substituting,
 \(\cfrac { { I }_{ max } }{ { I }_{ min } } =\cfrac { \left( 8 \right) ^{ 2 } }{ \left( 2 \right) ^{ 2 } } =\cfrac { 64 }{ 4 } 16\) (or)
\({ I }_{ max }:{ I }_{ min }=16:1\)

The given situation is shown in the figure
Given, A = 60^o ; i₁ = 0^o; i₂ = 90^o
(a) Equation for angle of deviation,
d = i₁ + i₂ - A
Substituting the values
d = 0^o + 90^o- 60^o = 30^o
The angle of deviation produced is, d = 30°
(b) The light inside the prism must be falling on the second face at critical angle as it graces the boundary. i_c = 90° – 30° = 60°
Equation for critical angle is, \(\sin i_{c}=\frac{1}{n}\) 
\(n=\frac{1}{\sin i_{c}} ; \quad n=\frac{1}{\sin 60^{\circ}}=\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}}=1.15\) 
The refractive index of the material of the prism is, n = 1.15

For a biconvex lens, radius of curvature of the first surface is positive and that of the second surface is negative as shown in the figure.

Given, n = 1.5, R₁ = 20 cm and R₂ = –15 cm
(a) Lensmaker’s formula \(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) 
Substituting the values,
\(\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-15}\right)=(1.5-1)\left(\frac{1}{20}+\frac{1}{15}\right)\) 
\(\frac{1}{f}=(0.5)\left(\frac{1}{20}+\frac{1}{15}\right)=(0.5)\left(\frac{3+4}{60}\right)=\left(\frac{1}{2} \times \frac{7}{60}\right)=\frac{7}{120}\) 
\(f=\frac{120}{7}\) = 17.14 cm
As the focal length is positive the lens is a converging lens. 
(b) When the lens is flipped by the side,

Now, R₁= 15 cm and R₂ = –20 cm, n = 1.5
Substituting the values in the lens maker's formula,
\(\cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ 15 } -\cfrac { 1 }{-20 } \right) \) 
\(\cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ 15 } +\cfrac { 1 }{ 20 } \right) \) 
This will also result in, f = 17.14 cm
Thus, it is concluded that the focal length of the lens will not change if it is flipped by the side. This is true for any lens. Students can verify this for any kind of lens.

Given, n = 4/3, d = 10 m.
Radius of illumination, \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \) 
\(R=\cfrac { 10 }{ \sqrt { \left( 4/3 \right) ^{ 2 }- } 1 } =\cfrac { 10\times 3 }{ \sqrt { 16-9 } } \) 
\(R=\cfrac { 30 }{ \sqrt { 7 } } =11.32cm\) 
To find the critical angle,
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ n } \right) \) 
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ 4/3 } \right) ={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) =48.6^{ o }\)
The total angle of view of the cone is, \({ 2i }_{ c }=2\times { 48.6 }^{ 0 }={ 97.2 }^{ 0 }\)