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Diktat Fisika X-2
Damriani
Dra. Damriani SMAN 3 Bandar Lampung 2008
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Diktat Fisika X-2
Damriani
SURAT KETERANGAN Nomor:
Yang bertanda tangan di bawah ini Kepala SMAN 3 Bandar Lampung menerangkan bahwa buku Diktat Fisika X-2 adalah benar ditulis oleh: Nama NIP Guru Guru Mata Mata Pelajar Pelajaran an
: Dra. Damriani : 131658096 : Fisika Fisika
dan telah digunakan sebagai material pembelajaran di SMAN 3 Bandar Lampung. Demikian surat keterangan ini dibuat untuk digunakan seperlunya.
Bandar Lampung, 4 Mei 2008 Kepala SMAN 3 Bandar Lampung
Drs. H E R N A D I NIP. 131870646
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Diktat Fisika X-2
Damriani
KATA PENGANTAR
Buku Diktat Fisika X-2 ini ditulis dengan rmaksud untuk digunakan oleh para siswa agar mudah mengerti pokok-pokok fisika dengan penjelasan yang ringkas dan sederhana dalam konsep, asas, dan prinsip fisika merupakan hal pertama yang harus dimengerti oleh para siswa. Dalam memecahkan soal-soal fisika, buku ini dapat digunakan untuk memberi gambaran global dari konsep-konsep fisika dan soal latihan yang terdapat di setiap bagian belakang bab dapat digunakan sebagai sarana melatih kemampuan memecahkan soal-soal fisika. Dengan selesai penulisan buku ini penulis mengucapkan t erima kasih kepada Drs. Hernadi sebagai Kepala SMAN 3 Bandar Lampung, atas semua dukungannya, masukan dan saran dari para kolega diucapkan terima kasih. Mereka adalah guru-guru fisika SMAN 3 Bandar Lampung, yaitu Zainal Abidin, S.Pd, Arif Santoso, S.Pd, Euis Waliah, Waliah, S.Pd, Dra. Sartinem dan Fera Nofrizawati, S.Pd. Buku ini tentu jauh dari sempurna, masukan, kritik dan saran yang membangun dapat disampaikan melalui email:
[email protected] [email protected].. Semoga kehadiran buku ini dapat memenuhi tujuan penulisan dan bermanfaat bagi penggunanya.
Bandar Lampung, 30 April 2008
Damriani
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Diktat Fisika X-2
Damriani
DAFTAR ISI
5. 6. 7. 8.
Surat Keterangan Kata Pengantar Daftar Isi
1 2 3
Alat-alat Optik Suhu dan Kalor Arus Listrik Gelombang Elektromagnetik
4 17 29 47
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Diktat Fisika X-2
Damriani
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Diktat Fisika X-2
Damriani
ALAT–ALAT OPTIK
MATA Kegunaan Kegunaan dari peralatan optik adalah untuk meperoleh penglihatan yang lebih baik, karena mata dapat dipandang sebagai alat optik maka pembahasan kita tentang alat optik kita mulai dari MATA sebagai ALAT OPTIK. Gambar
Bagian – bagian Mata: •
Bagian mata depan diliputi oleh membran transparan (C), yang disebut CORNEA.
•
Daerah di sebelah cornea mengandung cairan A, yang disebut : Aqueous Humor
•
Lebih ke dalam lagi, adalah lensa kristal (crystallinelensa) L, yang terdiri dari serat-serat yang cukup keras di pusat dan agak lunak di sebelah luar.
•
Lensa kristal ini terletak pada tempatnya, diikat dengan tali pada Ciliary Muscle (M).
•
Di samping lensa ini, mata dipenuhi cairan tipis (V), yang sebagian besar terdiri dari air,
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Diktat Fisika X-2
Damriani
penglihatan, disebut : Bintik buta. Bayangan yang dibentuk oleh mata terletak Retina, yaitu pada bagian yang disebut : Fovea centralis; urat-urat mata akan mengatur mata sedemikian rupa sehingga bayangan senantiasa jatuh pada fovea. Di depan lensa kristal, ada iris, terdapat ‘pembuka’ (P), yang disebut Pupil. Fungsi Fungsi pupil adalah untuk mengatur kuantitas sinar (intensitas cahaya) yang masuk ke mata. Pupil secara otomatis akan membesar jika intensitas cahaya rendah, sebaliknya akan berkontraksi berkontraksi jika intensitasnya jika intensitas cahaya bertambah. Proses semacam ini dikenal dengan : Adaptasi. Tetapi, Tetapi, secara relatif “besarnya” “besarnya” variasi daripada cahaya yang masuk ke dalam mata, tidak bisa hanya dikompensasi oleh perubahan ukuran pupil, oleh sebab itu adaptasi dalam retina sendiri yang bisa mengatasi perbedaan yang besar dalam kuantitas cahaya tersebut.
Beberapa istilah yang perlu diketahui pada mata: 1.
Daya Akomoda Akomodasi si
:
Daya menebal menebal dan menipisn menipisnya ya
lensa mata, lensa paling tipis pada saat mata tidak berakomodasi. 2.
Titik Jauh (Punctur Remotum
:
Titik
terjauh yang masih terlihat jelas oleh mata (tidak berakomodasi). Untuk mata normal : titik jauh letaknya di jauh tak berhingga. berhingga. 3.
Titik Titik Dekat Dekat (Punctur (Punctur Proxim Proximum) um)
: Titik Titik terdekat terdekat yang masih masih terlihat terlihat jelas oleh mata (berakomodasi (berakomodasi max )
Untuk mata normal : titik dekat 25 cm.
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Diktat Fisika X-2
Damriani
2.
Mata Mata Rab Rabun un Jauh Jauh (Myo (Myopi pi)) • • •
Mata tidak mampu melihat benda-benda jauh. Titik jauh mata lebih dekat daripada tah terhingga. Bayangan jatuh di depan retina, disebabkan karena :
-
Lensa mata terlalu cembung.
-
Lensa mata tidak dapat berakomodasi maksimum.
-
As mata (sumbu mata) terlalu panjang.
Gambar
Supaya dapat melihat seperti orang normal maka orang itu perlu bantuan kacamata lensa negatif (supaya sinar-sinar lebih divergen). 3.
Mata Mata Rabu Rabun n Dekat Dekat (Hyper (Hypermet metro ropi) pi) •
Mata tidak mampu melihat benda-benda dekat.
•
Titik dekat lebih jauh dari 25 cm.
•
Titik jauh dianggap tetap tak terhingga.
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1 1 1 + f =1 s=' = −1x 1 s + s ' = f Damriani s s ' f
Diktat Fisika X-2
positif (supaya sinar-sinar lebih divergen).
4.
Presbiopi Adalah kelainan mata pada orang tua, hal ini disebabkan : daya akomodasi berkurang. Dapat ditolong dengan kacamata lensa rangkap.
KACAMATA Kacamata pada dasarnya adalah sebuah lensa yang dipakai untuk mengatasi cacat mata, supaya diperoleh bayangan yang tepat dan jelas pada retina. Ada dua macam :
a.
Kacamata lensa positif b.
Kacamata lensa negatif
Bayangan yang dibentuk oleh kacamata senantiasa maya. (s’ dalam persamaan lensa tipis atau persamaan Gauss, senantiasa negatif).
Kacamata Lensa Positif Kacamata ini digunakan untuk mengatasi cacat mata rabun dekat (hypermetropi). Contoh: Seseorang yang titik dekatnya 75 cm (rabun dekat) ingin melihat sebuah benda yang letaknya 25 cm di depan mata (seperti mata normal), maka dia harus dibantu dengan kacamata lensa positif. Fungsi Fungsi dari kacamata kacamata ini adalah adalah untuk untuk membentuk membentuk bayangan bayangan dari sebuah benda benda (dalam (dalam contoh di atas s = 25 cm) supaya terletak pada titik dekatnya (dalam contoh di atas = 75 cm). Jadi dalam hal ini untuk kacamata berlaku : Untuk
:
s = 25 cm s’ = -75 cm (selalu maya)
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γ
β α
=
Diktat Fisika X-2
Damriani
s’ = - sd sd = jarak baca normal atau titik dekat mata
Adalah merupakan alat optik yang paling sederhana, hanya mempergunakan sebuah lensa cembung (positif). Letak : antara mata dan benda
Bayangan yang terbentuk : -
Maya
-
Diperbesar
Tegak Guna : Untuk melihat benda-benda kecil sehingga tampak lebih besar dan lebih jelas.
Gambar untuk mata berakomodasi berakomodasi
Gambar untuk mata tak berakomodasi berakomodasi
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D = − s '+ d Diktat Fisika X-2
sd s sd sd ' =× P = + P 1 ⋅ sd sd sd d f f D γ = + s −
f
Damriani
Keterang Keterangan an
D
D ⋅ f
: γ = perbesaran perbesaran anguler anguler β = sudut penglihatan tanpa loupe α = sudut penglihatan dengan loupe
Perbesaran Linier 1.
Untuk mata tak berakomodasi :
2.
Untuk mata berakomodasi
3.
Jika mata berjarak d dari lensa :
Perbesaran angulernya
:
:
dimana : Catata Catatan n
: Kalau Kalau di dalam dalam soal tentan tentang g loupe loupe tidak tidak dikatak dikatakan an apa-ap apa-apa, a, maka maka yang dimaks dimaksud ud
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s
f
=
ok ok s 'oby d s 'oby + f ok s 'oby sd sd = sd P = − × P ok = × + 1 P = − P P × soby sok = f Damriani P oby ok soby f ok ok Diktat Fisika X-2
P
s 'oby
= −
soby
sd
×
f ok
Jika d adalah jarak antara lensa obyektif dan okuler, maka :
Perbesarannya :
2.
Untuk mata berakomodasi berakomodasi maksimum Bayangan jatuh pada titik dekat dari pengamat.
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d = f + f f oby f sd f ok +ok oby oby P P = = f ok sd Damriani f
s
atau
s oby sd '
P P = = × sok s sok
Diktat Fisika X-2
oby
ok
ok
-
'
lensa okuler
• Benda letak jauh tak berhingga, sehingga bayangan jatuh pada fokus obyektif. • Fokus obyektif berikpit dengan fokus okuler. GAMBAR
d = jarak obyektif dan okuler
Perbesaran linier • Untuk mata tidak berakomodasi berakomodasi atau
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∞ sd sd oby = s='oby− + f 4oby f lp + f ok f s ' P oby ok = d f = oby + P f ok sok = f sd S ok P ok = Damrianiok sok =
=
Diktat Fisika X-2
f oby f oby sd + f ok f ok s sd oby P × M = '
= − × P f ok f ok sd sok sok =
Jarak obyektif okuler f lp = fokus lensa pembalik • Untuk mata tidak berakomodasi berakomodasi
• Untuk mata berakomodasi berakomodasi
c.
Teropon eropong g Pang Panggun gung g (Teropo eropong ng Bela Belanda nda = Ter Teropo opong ng Tonil onil = Teropo eropong ng Gali Galilei lei)) Terop Teropong ong panggung panggung dibuat dibuat sebagai sebagai pembaharu pembaharuan an dari teropong teropong bumi (karena (karena teralau teralau panjang). Untuk Untuk itu dipakai lensa lensa negatif (-) berfungsi sebagai sebagai lensa pembalik sekaligus sekaligus sebagai okuler.
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Diktat Fisika X-2
Damriani
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Diktat Fisika X-2
Damriani
SOAL LATIHAN
1.
Untuk Untuk mengama mengamati ti benda-b benda-benda enda renik renik digunak digunakan an ………… …………
2.
Untuk Untuk mengamati mengamati benda benda-bend -benda a yang jauh jauh letakny letaknya a digunaka digunakan n …………
3.
Untuk Untuk mengamati mengamati benda-b benda-benda enda kecil kecil agar tampak tampak lebih lebih besar dan dan lebih jelas jelas digunaka digunakan n… ………
4.
Sebuah Sebuah lup membentu membentuk k bayangan bayangan ……… yang yang jaraknya jaraknya dari dari mata paling paling sedikit sedikit ……… dan paling besar ………
5.
Perbesar Perbesaran an sebuah sebuah lup terletak terletak antara antara
dan
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Diktat Fisika X-2
Damriani
harus dipakai supaya dapat melihat benda-benda yang sangat jauh ? 6.a. Di mana titik dekat dekat sebuah mata mata yang memakai memakai kacamata baca baca 2 dioptri ? b.
Di mana titik titik jauh jauh sebuah sebuah mata mata yang yang memak memakai ai kacamat kacamata a -0,5 -0,5 dioptri dioptri untuk untuk meliha melihatt jauh ?
7.Sebuah mikroskop mempunyai obyektif yang berjarak titik api 10 mm dan okuler yang berj berjar arak ak titi titik k api api 25 mm mm.. Bera Berapa paka kah h jara jarak k anta antara ra kedu kedua a lens lensa a itu itu dan dan bera berapa paka kah h perbesaran perbesarannya nya apabila bendanya bendanya berada berada pada jarak 10,5 mm dari obyektif obyektif dan mata berakomodasi berakomodasi maksimum ? 8.Obyektif dan okuler sebuah mikroskop masing-masing mempunyai jarak titik api 2 cm.
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Diktat Fisika X-2
Damriani
a.
Jarak obyektif obyektif – okuler okuler untuk untuk mata mata tak berakomod berakomodasi asi
b.
Jarak obyekt obyektif if – okuler okuler untuk untuk mata yang yang berak berakomod omodasi asi pada pada jarak 20 20 cm
c.
Jarak Jarak dan jurusa jurusan n mengis mengisarn arnya ya okuler okuler untuk bayang bayangan an yang terang terang pada sebidan sebidang g tabir yang jaraknya 15 cm di belakang okuler.
d.
Lukislah Lukislah untuk untuk pertany pertanyaan aan b dengan dengan skala skala 1 : 5
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Diktat Fisika X-2
Damriani
13.Sebuah 13.Sebuah mikrosko mikroskop p mempunya mempunyaii onyektif onyektif yang berjarak titik api 7,5 mm. Benda kecil kecil berada 8 cm dari obyektif. Bayangan yang terbentuk dilihat dengan okuler yang berjarak titik api 5 cm. Pertanyaan : a.
Mata melihat melihat bayangan bayangan terang terang tanpa tanpa berakomod berakomodasi. asi. Berapa Berapa jarak obyektif obyektif dan okuler okuler..
b. Mata Mata berp berpen engl glih ihat atan an dek dekat deng dengan an titi titik k jauh jauh 20 cm dan dan meli meliha hatt baya bayang ngan an tak tak berakomodasi. berakomodasi. Berapa cm okuler harus digeser dan ke mana arahnya ? c.
Lukis Lukis pemben pembentuk tukan an baya bayanga ngan n pada pada b. b.
d.
Mata Mata berpen berpengli glihat hatan an dekat tadi mengula mengulang ngii penili penilika kann nnya ya sepert sepertii hanya hanya pada pada ad. a dengan menggunakan menggunakan kacamata sehingga okuler tdak harus digeser. digeser. Bila dalam hal ini
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Diktat Fisika X-2
Damriani
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200040000002d010000050000001402 200040000002d010000050000001402c0014000050000001302c0011c0115000000fb0280fe0000000000009001000 402c0014000050000001302c001280115000000fb0280fe0000000000009001000000000402001054696d6502004f0 00050000001402c0014000050000001302c001280115000000fb0280fe000000000000900100000000040200105469 C : R : (F-32) = 5 :0 4 : c0014000050000001302c0011c0115000000fb0280fe0000000000009001000 9 T = t C + 273 Diktat Fisika X-2
Damriani
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Diktat Fisika X-2
Damriani
L = Lo . . t
a. Termome ermometer ter alkoho alkohol. l.
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V AVt == AoVo. (1 t. t+ t) Diktat Fisika X-2
Damriani
At = Ao (1 + t)
t = tt1 – =tt01 – t0
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00d00009000000fa02000010000000000000002200040000002d010000050000001402000240000500000013020002b00115000000fb0280fe00000 00fb0280f ee0000000000009001000000000402001054696d6502004f006c006500500072006500730030003000300000000000000000000000000000 0000000000009001000000000402001054696d6502004f006c006500500072006500730030003000300000000000000000000000000000 Q=H . t LUSSAC1 joule = 0,24 kal BOYLE – GAY 1 kalori = 4,2 joule 0000000000009001000000000402001054696d6502004f006c00650050007200650073003000300030000000000000 000000000 0009001000000000402001054696d6502004f006c00650050007200650073003000300030000000000000 00000fa02000010000000000000002200040000002d010000050000001402000240000500000013020002d00115000000fb0280fe000000000000900 000000000009001000000000402001054696d6502004f006c0065005000720065007300300030003000000000000000000000000000000000000000 Diktat Fisika X-2
Damriani
Atau
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Diktat Fisika X-2
Damriani
= 2.
29.Hitunglah hambatan pengganti dari rangkaian di bawah ini.
30.Dari rangkaian di bawah ini, maka tentukan arus yang dihasilkan Baterai.
31.Hitunglah arus yang dihasilkan baterai pada rangkaian yang dibawah ini.
32. Tentukan Tentukan arus yang dihasilkan baterai pada rangkaian di bawah ini.
33. Tahanan Tahanan PA = BN = R. a. Hitung arus yang melalui cabang ADB dan ACB. b. Hitung beda potensial antara A dan B c. Hitung berapakah tahanan PA.
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Diktat Fisika X-2
Damriani
34.Hitunglah V ab
35.Untuk rangkaian di bawah ini jika S 1 dan S2 ditutup, maka voltmeter ( V ) akan menunjukkan harga............
36.Dua batang kawat terbuat dari perak dan platina dihubungkan dihubungkan secara seri. Kawat perak panjangnya 2 meter, penampangnya 0,5 mm 2, hambatan hambatan jenisnya jenisnya 1,6.10-8 ohm meter. Sedangkan kawat platina panjang 0,48 m. Penampangnya 0,1 mm 2 dan hambatan jenisnya 4.10 -8ohm meter. Hitung berapa kalori panas yang timbul pada kawat platina, jika ujungujung rangkaian tersebut diberi tegangan 12 volt selama 1 menit.
37. Jika Jika di ketahui : r 1 = 0,5 ohm ; R 1 = 1,5 ohm ; r 2 = 1 ohm ; R2 = 2 ohm ; E 1 = 2 V ; 1 V ; E3 = 1,5 V ; E4 = 2,5 V ; R 5 = 2 ohm ; r 3 = 0,5 ohm ; R 3 = 1 ohm ; R4 = 2 ohm. Hitunglah I 1, I2 dan I3.
E2 =
r 4 = 1 ohm
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Diktat Fisika X-2
Damriani
38.Pada gambar di samping. Hitunglah besar tentukan arah dari I 1, I2 dan I3 ?
39.Sebuah bujursangkar bujursangkar ABCD dibuat dari kawat yang berbeda-beda, tahanan
AB =
2 ohm, tahanan BC = 7 ohm, tahanan CD = 1 ohm. Tahanan DA = 10 ohm sedangkan diagonal BD dihubungkan dengan tahanan dari 2 ohm. Titik A dihubungkan dengan Kutub + dari elemen baterai yang tahanan dalamnya 1 ohm sedangkan titik C dihubungkan dengan kutub - dari elemen tersebut. Kuat arus induk dari kutub + elemen yang masuk ke titik A adalah 1 Ampere. a. Berapa besar dan arah arus yang melalui diagonal BD. b. Berapa besar dan arah arus yang lain pada setiap cabang. c. Berapakah GGL elemen tersebut.
40. Titik Titik p, A, dan N terletak pada satu garis lurus. Tahanan PA = tahanan BN. Beda potensial antara titik P dan N = 12 V. Antara A dan B terdapat 2 cabang yaitu ADB dan ACB kuat arus yang melalui PA PA = 5 ampere. Tahanan AD = 1 ohm tahanan DB = 2 ohm, tahanan CB = 3 ohm. Sedangkan antara A dan C terdapat sebuah elemen kutub + dihubungkan titik A, Kutub - dengan titik C. GGL elemen 2 V. Tahanan dalamnya 0,5 ohm.
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Diktat Fisika X-2
Damriani
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Diktat Fisika X-2
Damriani
GELOMBANG
ELEKTROMAGNETIK
Bila dalam kawat P Q terjadi perubahan-perubahan tegangan baik besar maupun arahnya, maka dalam kawat PQ elektron bergerak bolak-balik, dengan kata lain dalam kawat PQ terjadi getaran listrik. Perubahan tegangan menimbulkan perubahan medan listrik dalam ruangan disekitar kawat, sedangkan perubahan arus listrik menimbulkan perubahan medan magnet. Perubahan medan
listrik dan medan magnet itu merambat merambat ke segala jurusan. jurusan. Karena rambatan rambatan perubahan medan magnet dan medan listrik secara periodik maka rambatan perubahan medan listrik dan medan magnet lazim disebut : GELOMBANG ELEKTROMAGNETIK. Percobaan-percobaan Percobaan-percobaan yang teliti membawa kita pada kesimpulan : 1.
Pola Pola gelombang gelombang elektromag elektromagnetik netik sama sama dengan pola pola gelombang gelombang transvers transversal al dengan vektor perubahan medan listrik tegak lurus pada vektor perubahan medan magnet.
2.
Gelombang Gelombang elektromag elektromagnetik netik menunjuk menunjukkan kan gejala-gej gejala-gejala ala : Pemantulan, Pemantulan, pembiasan, difraksi, polarisasi seperti halnya pada cahaya.
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Diktat Fisika X-2
Damriani
Hukum Coulomb
:
Hukum Biot-Savart
:
Hukum Fa Faraday
:
Muatan lilistrik me menghasilkan medan listrik yang kuat. Aliran muatan (arus) listrik menghasilkan medan magnet disekitarnya. Perubahan me medan ma magnet (B (B) da dapat me menimbulkan me medan li listrik (E).
Didorong oleh keyakinan atas keteraturan dan kerapian hukum-hukum alam, Maxwell berpendapat : Masih ada kekurangan satu aturan kelistrikan yang masih belum terungkap secara empirik. Jika perubahan medan magnet dapat menimbulkan perubahan medan listrik maka perubahan medan listrik pasti dapat menimbulkan perubahan medan magnet, demikianlah keyakinan Maxwell. Dengan pengetahuan matematika yang dimilikinya, secara cermat Maxwell membangun teori yang dikenal sebagai teori gelombang elektromagnetik. Baru setelah bertahun-tahun Maxwell tiada, teorinya dapat diuji kebenarannya kebenarannya melalui percobaan-percobaan.Menuru percobaan-percobaan.Menurutt perhitungan secara teoritik, kecepatan gelombang elektromagnetik hanya bergantung pada permitivitas (
ε 0 ) dan permeabilitas ( µ 0 ). 1
c=
. 0 ε 0 µ 1
.10−9
µ 0
−7
Dengan memasukkan ε 0 = 4π .9 C/N.m2 dan = 4π .10 W/A.m 8 Diperoleh nilai c = 3.10 m/s, nilai yang sama dengan kecepatan cahaya. Oleh sebab itu Maxwell mempunyai cukup alasan untuk menganggap cahaya adalah Gelombang Elektromagnetik. Elektromagnetik. Oleh karena itu konsep gelombang elektromagnetik ini merupakan penyokong teori HUYGENS tentang cahaya sebagai gerak gelombang. INTENSITAS INTENSITAS GELOMBANG ELEKTROMAGNETIK
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Diktat Fisika X-2
Damriani
Kedua medan listrik dan medan magnet tersebut saling tegak lurus, merambat kearah sumbu X. Kedua gelombang tersebut dapat dituliskan menjadi : Ey = E0 sin (kx- ω t) Ez = B0 sin (kx- ω t) Intensitas gelombang elektromagnetik dituliskan menjadi :
E y.B z s=
µ 0
E 0.B0 s=
µ 0
sin2 (kx- ω t)
Jadi hanya intesitas (s) tergantung dari sin 2 (kx- ω t), s akan berharga maksimum bila harga sin2 (kx- ω t) = 1, atau
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Diktat Fisika X-2
Damriani
s=
Emax
+ Bmax
2 µ 0
Selain itu
s=
1 2
s juga dapat dituliskan menjadi :
ε 0 E02 c
Karena :
1) E0 = c B0 ; E0 = Emax dan B0 = Bmax
1 2) c = Nilai
µ 0ε 0
s juga dapat dituliskan dalam bentuk : 2
s
=
E0 2c µ 0
Gejala gelombang elektromagnetik baru dapat ditunjukkan beberapa tahun setelah Maxwell meninggal oleh : H.R. Hertz.
Spektrum Gelombang Elektromagnetik Menurut panjang gelombang ( λ ) dan frekuensi (f) dapat disusun spektrum gelombang elektromagnetik sebagai berikut : F(Hz)
λ(Ao)
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Diktat Fisika X-2
Damriani
CAHAYA Gelombang 10 -5 Infra merah 10
10
Radar &
5
10
Televisi Gelombang
105
Radio 100
Dari spektrum tersebut dapat disimpulkan disimpulkan bahwa makin pendek panjang gelombang ( makin tinggi fekuensinya fekuensinya (f) dan makin besar pula pula daya tembusnya.
λ )
Diantara gelombang-gelombang yang terdapat pada spektrum tersebut, yang dapat dilihat o
oleh mata hanyalah gelombang cahaya yang mempunyai panjang gelombang antara 8000
A
o
(merah) - 4000 A (ungu). Gelombang yang mempunyai daya tembus yang sangat besar adalah sinar X dan sinar
γ . γ
Dimana sinar X dihasilkan dengan cara EMISITHERMIONIK , sedangkan sinar dihasilkan oleh inti-inti yang tidak stabil (bersifat radioaktif). Manfaat gelombang elektromagnet elektromagnet dapat diterangkan sesuai urutan spektrumnya :
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Diktat Fisika X-2
Damriani
c x ∆t s=
2
c = kecepatan cahaya (3 x 10 8 m/det) 4. Daer Daerah ah fre frek kuens uensii 1011 – 1014 Hz, ditempati oleh radiasi infra merah, dimana gelombang ini lebih panjang dari gelombang cahaya tampak dan tidak banyak dihamburkan dihamburkan oleh partikel-partikel partikel-partikel debu dalam atmosfir sehingga mengurangi batas penglihatan manusia. 5. Daer Daerah ah fre frek kuens uensii 1014 – 1015 Hz, berisi daerah cahaya tampak (visible light), yaitu cahaya yang tampak oleh mata manusia dan terdiri dari deretan warna-warna merah sampai ungu. 6. Daer Daerah ah fre frek kuens uensii 1015 – 1016 Hz, dinamakan daerah ultra ungu (ultra violet). Dengan frekuensi ultra ungu memungkinkan kita mengenal lebih cepat dan tepat unsur-unsur yang terkandung dalam suatu bahan. 7. Daer Daerah ah fre frek kuens uensii 1016 – 1020 Hz, disebut daerah sinar X. Gelombang ini dapat juga dihasilkan dengan menembakkan elektron dalam tabung hampa pada kepingan logam. Karena panjang gelombangnya sangat pendek, maka gelombang ini mempunyai daya tembus yang cukup besar sehingga selain digunakan di rumah sakit, banyak pula digunakan di lembaga-lembaga penelitian ataupun industri. 8. Daer Daerah ah fre frek kuens uensii 1020 – 1025 Hz, disebut daerah sinar gamma. Gelombang ini mempunyai daya tembus yang lebih besar daripada sinar X, dan dihasilkan oleh intiinti atom yang tidak stabil.
Radiasi Kalor. Benda-benda yang dipanasi mengemisikan gelombang yang tidak nampak (sinar ultra ungu dan infra merah). Radiasi dari benda-benda yang dipanasi disebut radiasi kalor. kalor. Benda-benda yang dapat menyerap seluruh radiasi yang datang disebut benda hitam mutlak, sebuah kotak yang mempunyai lubang sempit dapat dianggap sebagai benda yang hitam mutlak.
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Diktat Fisika X-2
Damriani
Ketergantungan kekuatan radiasi suatu benda terhadap panjang gelombangnya disebut spektrum radiasi (spektrum gelombang pancaran). Eksperimen-eksperimen Eksperimen-eksperim en untuk mengamati spektrum radiasi telah dilakukan, hasil spektrum radiasi carbon pada berbagai suhu seperti terlukis pada gambar. Dari diagram itu Wien mengambil kesimpulan yang dikenal sebagai : HUKUM WIEN.
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Diktat Fisika X-2
Damriani
o
a.
8000
A
6000
A
o
b. 4.
Pada Pada suhu berapa berapa derajat derajat Celcius Celcius energi energi radiasi radiasi maksimum maksimum benda benda hitam hitam sempurna sempurna o
berada pada sinar yang panjang gelombangnya 2000
A
5.
Sebuah Sebuah lampu lampu busur busur memancark memancarkan an cahaya cahaya pada suhu 3000 0C. a. Berapa Berapa panjang panjang gelombang gelombang sinar sinar yang yang energi energi radiasi radiasinya nya maksimum. b. Berapa Berapa besar besar energi energi radia radiasi si tiap tiap detik detik tiap tiap satuan satuan luas. luas.
6.
Sepotong Sepotong platina platina menjad menjadii merah merah pijar pijar pada pada suhu suhu 550 0C dan putih pijar pada suhu 0 1600 C. Berapakah panjang gelombang yang energi radiasinya maksimum pada suhu-suhu suhu-suhu tersebut.
7.
Sebuah Sebuah gelomb gelombang ang radi radio o mempu mempunya nyaii E m = 10-4 V/m Berapakah besar medan magnet
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Diktat Fisika X-2
Damriani