Homework 1. Property LASER Incandescent Bulb

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1 Homework 1 Solution: a) LASER light is spectrally pure, single wavelength, and they are coherent, i.e. all the photons are in phase. As a result, the beam of a laser light tends to stay as beam, and not diverge due to scattering. It works on the principle of stimulated transitions between the electron energy levels and is also highly amplified. Incandescent bulb works by black body radiation. A tiny filament is heated up to extremely high temperatures and as heated elements radiate light, the bulb glows. The light emitted in this process has a whole range of energies (and wavelengths) and has no preferential direction of travel or polarization or phase. This light is not amplified. The differences can be summarized as follows: Property LASER Incandescent Bulb Nature of emission Stimulated emission Spontaneous Emission Coherence Coherent Incoherent Directionality Highly directional (focused to a very small point) Divergent (cannot be focused to a small point) Monochromatic Polychromatic Amplified Not amplified b) The following ingredients must be present to make a laser to work: (i) Lasing / Active / Gain medium (ii) Optical Cavity (iii) Resonator (iv) External energy source (Pump) Lasing medium is excited by the external energy source (pump) to produce population inversion. Spontaneous and stimulated emission of photons takes place, leading to the phenomenon of optical gain (amplification). Common medium include Ruby, He- Ne, YAG, etc. Pump provides energy required for the population inversion and stimulated emission to the entire system. Either electrical discharge or optical discharge can be used as pumping sources. Resonator guides the light about the simulated emission process induced by high-speed photons. There is also a fully reflective and a partially reflective mirror. Both are set up on optical axis, parallel to each other. The gain medium is located in the optical cavity between the two mirrors. This setup makes sure that only those photons which came along the axis, pass and others are reflected by the mirrors back into the medium, where it may be amplified by stimulated emission. 1

2 Solution: Sketch From the sketch, D is the diameter of the optical beam; L and R is the length and the radius of the resultant cone, respectively. θ is the angle of divergence of the beam. R = D + x = D + L tan θ D + Lθ (1) (small angles tan x = x) From Uncertainty relationships and spread of the Gaussian beam, we know that: θ = k y k z = λ π( D ) = λ πd () From (), (1) becomes, Volume of the resultant cone becomes, R = D + Lλ πd V = 1 3 πr L = 1 3 πl [(D ) + ( RD ) + R ] (3)

3 Function to be minimized is V ( D ). Putting value of R in (3) and differentiating gives, L [ D L ] 4 = 4 3 [λ πl ] 4 D = m 633nm { } π θ = λ πd = rad Resultant cone diameter = R = ( D + Lθ ) = m = m 4 Answer: Minimum beam diameter Expression is D = 4 3 { L λ π } and the value for the given values is D = m Solution: (a) From the equations above, For TEM 0,0 case, x = x E(x) dx E(x) dx (1) k x = k x E(k x ) dk x E(k x ) dk x () E(x) = E 0 exp [ ( x ) ] (4) 3

4 Putting this is in equation (1), and integrating by method of substitution, we get, Let t = x x (E 0 exp [ ( x x w ) ]) dx = 0 (E 0 exp [ ( x w ) ]) dx 0 dt dx =. Then (3) becomes, x = x x exp [ ] d ( x w ) w 3 0 /( 3 ) 0 exp [ x ] d ( x ) / Evaluating the above expression on Wolfram Alpha, we get, x = = (1 ) x = Expression for E(x) - (4) above in k-space can be written as, (3) ( t exp( t ) dt) exp( t ) dt E(k x ) = E 0 exp [ ( x ] w ) exp( jk x x) dx = π E 0 exp ( ( k x 0 ) ) Putting this expression in Eq (), we get, k x = k x [ π E 0 exp ( ( k x ) )] dk x [ π E 0 exp ( ( k x ) )] dk x (5) Let k t = k x k x = d(k t ) =. Then (5) becomes, dk x { w 0k x } exp [ k x ] d (k x ) (3 3 )/ exp [ k x ] d (k x ) / Evaluating the above expression on Wolfram Alpha, we get, k x = (1 ) k x = 1 = w {( k t exp( k t ) dk t 0 exp( k t ) dk t ) } (b) The given field is, E 10 = ( x ) exp [ x + y ] (6) 4

5 This can be written in k-space as, Let t = x + j k x z E 10 (k x ) = E 0 x = t j k x E 10 (k x ) = exp [ k x ] x exp ( [ x ] ) exp( jk w x x) dx 0 = j k k x x w o e ( π ) (7) Putting (6) in (1), and integrating by substitution, we get, ( exp( t ) t dt j k x exp( t ) x = t4 exp( t ) dt = t exp( t ) dt (3 ) x = 3 Putting (7) in (), and integrating by substitution, we get, k k w 4 t exp( k t ) dk t x = 0 = k t exp( k t ) dk t w (3 0 ) k x = 3 Uncertainty product is given by, (c) Sketch of intensity vs x: x k x = 3 3 = 3 dt) 5

6 Solution: (a) We are given that k H = ωd (1) Taking a dot product of that with k, we have, k (k H) = 0 () This is because cross product of vectors is normal to both the vectors. Now, putting () in (1), we get, k ( ωd) = 0 k D = 0 as required. ---(3) This is because, ω 0. (b) We are given that k H = ωd D = ( 1 ) (k H) (4) ω k E = ωb B = ( 1 ) (k E) (5) ω Using the following property of cross products, (A B) (C D) = C[A (B D)] D[A (B C)] From (4) & (5), we have D B = ( 1 ω ) {k[k (H E)] E[k (H k)]} (6) From (), the second term goes to 0 and k (H E) 0 and is a scalar Therefore, 6 becomes: 6

7 D B = ( 1 ω) [k (H E)] (k) (7) scalar equation (7) shows that wave vector, k and D B are in the same direction. (c) k (k E) = k (ωb) = ω(k (μ 0 H)) = ωμ 0 (k H) = ωμ 0 ( ωd) = ω μ 0 D (8) k (k E) = [k(k. E) k E] = ω μ 0 D (9) Taking dot product of both sides of (9) with D, we get D [k(k. E) k E] = ω μ 0 D. D D k(k. E) D k E = ω μ 0 D. D (10) D.k=0 (from a) k = ω μ 0 D. D D. E (d) In anisotropic medium, Poynting vector, S = ( 1 ) E H k E = ωb = ωμ 0 H S = ( 1 E k ) E ωμ 0 Let us take the case of a linear anisotropic medium. D is no longer parallel to E From the attached sketch it can be seen that D = 1 k (k E) = k [E (u E)u] μ 0 ω μ 0 ω D is the projection of E in the plane orthogonal to u (to within a mult. factor) Therefore the Poynting vector (direction of «light ray») is no longer parallel to the wave vector (direction of propagation of the phase). 7

8 Solution: Problem.1 gives us the following: Problem. gives us the following: Combining the two for the negative lens, we will have the following steps of propagation (from the sketch): (i) Propagation from medium of n 1 refractive index at spherical dielectric interface to a medium of index n (Result from problem.1) Ray part 1 (ii) Propagation in a dielectric(same medium) of refractive index n for a distance d - Ray part 8

9 (iii) Propagation from medium of n refractive index at plane dielectric interface to a medium of index n 1 ((Result from problem.) Ray part 3 Thus the ABCD matrix of the entire system can be written as: 1 0 A B [ C D ] = [ 0 n ] [ 1 d 1 0 n 0 1 ] [ (1 n 1 ) ( 1 1 d 1 n R ) n 1 ] = [ 0 n 1 0 ] [ (1 n 1 ) ( 1 n n 1 n R ) n 1 ] n ray part 3 = Verification: det(abcd) = 1 ray part ray part ( d R ) (1 n 1 dn 1 ) 1 + ( d n n R ) (1 n 1 ) dn 1 n n [ n 1 R (1 n = n 1 1 ) 1 n ] [ R (n 1) 1 n 1 ] The resultant ABCD matrix for the entire system is as follows: 1 + ( d A B [ C D ] = R ) (1 n 1 ) dn 1 n n 1 [ R (n 1) 1 n 1 ] Solution: (a) For a plane mirror, the ABCD matrix is A B [ C D ] = [ ] 9

10 We can see that this is similar to that of a spherical mirror with R. The plane mirrors can be considered just as directors of the optical axis and can be ignored in a unit cell for propagation as their ray matrix is just an identity matrix. The resultant equivalent lens arrangement is: (b) To obtain the ABCD matrix for the system, ( r 1 0 r ) = ( 1 d f 1) (1 ) ( 1 d d d ) ( ) (1 ) ( r r 1 ) = ( 1 3d 1) (1 0 1 ) ( r 1 r f 1 ) 1 3d = ( 1 f 1 (3d f )) ( r 1 r 1 ) 1 3d A B [ C D ] = [ 1 f 1 (3d f )] (c) Values of d that make the cavity stable: Stable if - f 1 < A + D < 1 3d f 1 < < 1 < 3d f < 4 < 3d f < < d f < 0 0 < d f < 4 3 Values of d f that make the cavity stable are d f (0, 4 3 ) 10

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