Izibonelo Zemibuzo Yengxoxo Ye-Microwave
Ama-microwave ayingxenye ye-spectrum ye-electromagnetic enama-wavelength asukela ku-1 mm kuya ku-1 m. Lawa magagasi avame ukusetshenziswa kwezobuchwepheshe obuhlukahlukene besimanje, njengokuxhumana okungenantambo, i-radar, i-astronomy, kanye nezinto zasekhaya ezifana nama-microwave oven. Njengoba kubhekwa ukubaluleka kokusetshenziswa kwama-microwave, kubalulekile kubafundi, ikakhulukazi labo abafunda i-physics noma ubunjiniyela, ukuqonda izisekelo zama-microwave, okuhlanganisa nendlela yokuxazulula izinkinga ezihlobene nawo. Lesi sihloko sizoxoxa ngezibonelo eziningana zezinkinga futhi sixoxe ngama-microwave.
Umbuzo 1: Ubude be-Wave kanye ne-Vacation
Umbuzo:
Njengoba i-microwave inemvamisa engu-10 GHz, bala ubude be-wavelength ye-microwave. (Sebenzisa isivinini sokukhanya c = 3 x 10^8 m/s)
Ingxoxo:
Okokuqala, sikhumbula ubudlelwano phakathi kobude be-wavelength (λ), imvamisa (f), kanye nesivinini sokukhanya (c):
\[ c = \lambda \cdot f \]
Singakwazi noma siphinde le fomula ukuthola ubude be-wavelength (λ):
\[ \lambda = \frac{c}{f} \]
Njengoba sinikezwe imvamisa (f) = 10 GHz = 10 x 10^9 Hz, kanye nesivinini sokukhanya (c) = 3 x 10^8 m/s, sixhuma la manani amabili kufomula:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{10 \times 10^9 \, \text{Hz}} \]
\[ \lambda = 0.03 \, \text{m} \]
Ngakho-ke, ubude be-wavelength ye-microwave buyi-0,03 metres noma i-3 cm.
Umbuzo 2: Amandla Okudlulisa kanye Nebanga
Umbuzo:
I-transmitter ye-microwave ikhipha amandla angu-50 W. Bala ubukhali be-microwave ebangeni elingamamitha amabili ukusuka emthonjeni we-transmitter, ucabangela ukusabalala kwe-isotropic.
Ingxoxo:
Ukuze sibale ukuqina, sisebenzisa i-equation yendawo yamandla ngeyunithi ngayinye:
\[ I = \frac{P}{A} \]
I-P ingamandla (50 W) kanti i-A yindawo engaphezulu yesikwele esinobubanzi obungamamitha amabili:
\[ A = 4 \pi r^2 \]
Faka inani r = 2 amamitha kufomula yendawo:
\[ A = 4 \pi (2 \, \text{m})^2 \]
\[ A = 16 \pi \, \text{m}^2 \]
Bese ubala ubukhali:
\[ I = \frac{50 \, \text{W}}{16 \pi \, \text{m}^2} \]
Ukubala amanani ezinombolo:
\[ I \cishe \frac{50}{50.265} \]
\[ I \cishe 0.995 \, \umbhalo{W/m}^2 \]
Ngakho-ke, amandla e-microwave ebangeni elingamamitha amabili ukusuka emthonjeni wokudlulisa angaba ngu-0.995 W/m².
Umbuzo 3: Umphumela we-Doppler kuma-Microwave
Umbuzo:
Imoto ehamba ngesivinini esingu-108 km/h (30 m/s) isondela kuma-microwave adlulisa i-radar ngesivinini esingu-5 GHz. Bala imvamisa etholwa yi-radar uma isivinini se-microwave siyisivinini sokukhanya.
Ingxoxo:
Sebenzisa umphumela we-Doppler wemvamisa etholiwe (f'):
\[ f' = f \left(\frac{c + v}{c}\right) \]
Di mana:
– f' = imvamisa etholiwe
– f = imvamisa yomthombo = 5 GHz = 5 x 10^9 Hz
– c = isivinini sokukhanya = 3 x 10^8 m/s
– v = isivinini semoto = 30 m/s
Faka amanani kufomula:
\[ f' = 5 \izikhathi ezingu-10^9 \kwesobunxele(\frac{3 \izikhathi ezingu-10^8 + 30}{3 \izikhathi ezingu-10^8}\kwesokudla) \]
\[ f' \cishe 5 \izikhathi 10^9 \kwesobunxele(1 + \frac{30}{3 \izikhathi 10^8}\kwesokudla) \]
\[ f' \cishe 5 \izikhathi 10^9 \kwesobunxele(1 + 1 \izikhathi 10^{-7}\kwesokudla) \]
\[ f' \cishe 5 \izikhathi 10^9 \izikhathi 1.0000001 \]
\[ f' \cishe 5.0000005 \izikhathi eziyi-10^9 \]
\[ f' \cishe 5.0000005 \, \umbhalo{GHz} \]
Ngakho-ke, imvamisa etholwa yi-radar icishe ibe yi-5.0000005 GHz.
Umbuzo 4: Ukumuncwa Kwamandla Ngezinto Ezisetshenziswayo
Umbuzo:
Into ethile imunca amandla e-microwave angu-2 W lapho amandla egagasi lesigameko engu-10 W/m². Bala indawo engaphezulu kwento.
Ingxoxo:
Sebenzisa ubudlelwano phakathi kwamandla okumunca nendawo engaphezulu kanye nokuqina:
\[ P = I \cdot A \]
Faka amanani aziwayo esikhundleni sawo:
\[ 2 \, \umbhalo{W} = 10 \, \umbhalo{W/m}^2 \umbhaloA \]
Ukuxazulula i-A:
\[ A = \frac{2 \, \text{W}}{10 \, \text{W/m}^2} \]
\[ A = 0.2 \, \umbhalo{m}^2 \]
Ngakho-ke, indawo engaphezulu yezinto ezibonakalayo ingu-0,2 m².
Umbuzo 5: Amaphethini Okuzindla Nokuphazamisa
Umbuzo:
Ama-antenna amabili e-microwave akhipha amaza obude obungu-6 cm ohlangothini olufanayo. Uma ibanga eliphakathi kwama-antenna lingu-12 cm, kunini lapho ukuphazamiseka okwakhayo kwenzeka khona kuqala?
Ingxoxo:
Ukuphazamiseka okwakhayo kwenzeka lapho umehluko wendlela phakathi kwemithombo emibili ungu-k = ±nλ (n iyinombolo engeyona eyero ehlanganisa u-0). Okokuqala ngqa ukuphazamiseka okwakhayo (n=1):
\[ d = n \lambda/2 \]
Njengoba ibanga le-antenna lingu-12 cm kanti ubude be-wavelength bungu-6 cm, umehluko wendlela yokuphazamiseka kokuqala kokwakha kwenzeka ku:
\[ d = n (\lambda/2) \]
Ngakho-ke:
\[ 12 = 1 \izikhathi (6/2) \]
Ngakho-ke ibanga elitholakalayo yileli:
\[ 12 = 1 \izikhathi 3 \]
Ngenxa yokuthi lesi sibalo sihlobene nokwakheka kwamaphethini okuphazamiseka kulokho okuvamile, khona-ke kusukela ekucabangeni siklama emgqeni ophakathi webanga ukusuka emthonjeni noma endleleni ngokusekelwe endaweni engeyona ithrekhi, okungukuthi indawo eqhubekayo ye-polar yomugqa womkhondo engahlobene nephethini yesikhundla esinqunyiwe esinqunyiwe.
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Lezi ezinye zezibonelo zezinkinga ze-microwave kanye nendlela yokuzixazulula. Ngokuqonda nokujwayela ukuxazulula izinkinga ezinjengalezi, kunethemba lokuthi kuzoba lula ngawe ukuqonda nokusebenzisa imiqondo ye-microwave emikhakheni ehlukahlukene yesayensi nobuchwepheshe.
