soal dan jawaban Gelombang & Optik

1. Seorang teknisi kedokteran gigi menggunakan cermin kecil yang memberikan perbesaran 4 kali bila dipegang 0,60 cm dari gigi. Berapakah jari-jari permukaan cermin?

           

 

diketahui nilai perbesaran: perbesaran: = 4 dan dan = 0,60 0,60



maka, nilai q adalah = 4 = 4 × 0,60 = 2,40 2,40 persamaan persamaan cermin menjadi: 1 1 2 + = 0,60 2,40 2,40 + 0,60 0,60 2 = 1,44 2 × 1,44 = 3,00 = 0,96 0,96 Jadi, jari-jari permukaan cermin adalah 0,96 cm bernilai positif sehingga cermin haruslah cekung.



2. Cahaya hijau ( =540 nm) didifraksi oleh kisi dengan 2000 garis/cm. a. hitung sudut difraksi orde ke-3!

−  λ ∙      −   λ ∙           −  −             −     − ∙   − −     −     − 3

3 × 5,4 5,4 10 sin 3 = = 1 2000 18,9 ,9°° 3 = 18 b. Adakah orde ke-10 yang mungkin terjadi?

5

= 0,32 0,3244

10 × 5,4 5,4 10 5 sin 10 = = = 1,08 1,08 1 2000 Karena nilai sin   tidak mungkin melebihi satu, maka orde ke-10 tidak mungkin terjadi. 10

3. Selidiki apakah

,

= 4+(2

0,1 10 )2

merupakan suatu suatu gelombang. Jika ya,

berapakah berapakah cepat rambat, frekuensi dan amplitudonya! ,

=

= 0,1 4 + 2 + 10

0,1 4 + 2 + 10 =2 2

10

2

2

2

22

4 + 2 + 10

1

10

2

2

10

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 −     − −     − ∙   − −   −     − −  −      −  −     − ∙   −    −   −     −  −      −   −  −     −    ∙  −   −   −   −  −    −    −   −              −     ∙ − 2

2

2

2

= 20 4 + 2 + 10

2

= 2(( 10) 4 + 2 + 10 2

2

=

+

2 (1

2

2

= 0,4 2 4 + 2 + 10 22

2

2

2

22

10

4 + 2 + 10

+ 2

22

2

10 2

10 )

1

2

2

10 ( 2) 4 + 2 + 10

2

3

10 2

2

2

10

2

3

10 )2 4 + 2 + 10

4(2

0,1 4 + 2 + 10

= 0,4 2

2

2 4 + 2 + 10

10 )2 1

= 0,8 4 + ( 2

4(2

10 )2 4 + ( 2

10 )2

2

2

2

2

=

20 200 = =25= 0,8 8

2

jadi ( , ) adalah fungsi gelombang. Maka = 5 / dengan x dalam meter dan t dalam detik. 5 = 1 0 = 2  atau = Amplitudo adalah simpangan terbesar, artinya ( , ) maksimum, jika penyebutnya minimum, atau (2 10 )2 = 0. 0,1 1 Jadi, 0 = 4 = 40 = 0,02 0,0255 = 25 10 2 4. Tentukan koefisien amplitudo (=koef. refleksi dan koef. transmisi) pada jatuh normal, dan hitunglah untuk n 2 = 1,5 dan n 1 = 1 (udara)!

∥  ′′ −−   ⊥ ′′ ′′  −     ′′   =

2 cos

 

+

′′  −  ′′′′  −  =

2

cos 1+ 1 cos 2 cos 2 = = 1 cos + 2 cos 1+ = 0 mak makaa = 0 ; cos cos = 1 ; cos cos = 1 (jatuh (jatuh norma normal) l) 1,5 1 0,5 1 Jadi =  pada jatuh normal = 1,5+1 = 2,5 = 5 = 20% 2 1 cos 2 1 = = + 2 cos 1 cos 1+ 2 2 1 cos 2 1 = = + 2 cos 1 cos 1+ 2 2 2 8 Jadi, =  pada jatuh normal = 1+1,5 = 2,5 = 10 = 80%

 ∥ −′⊥ ∥  ′ ⊥  ′ ∥ ⊥

1 cos

1 cos 2

1 2 1 2

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     −    ′ k f o kus bayangan  F ’  adal a h j a r a k s ’ ; j i k a s  ∼ ∝−  ′  −  ′     ′  , ′ jadi− letakt′itik −fokus bayangan  ′ −F’ adalah 40 cm di belakang lensa negatif. Let ′ ak∼titik fokus∝  benda  F ’  adal a h pada j a r a k s ’ = t a k hi n gga −  −   ′  −− −− −    ′′    −  ′′ − −−    ′  ′ −−   −      ′ 

5. Sebuah lensa gabungan terdiri dari lensa positif dan negatif yang berjarak 20 cm.  ; 2 = 40 . Tentukan letak titik fokus benda dan bayangan dari 1 = +40 lensa gabungan ini! Hitung  dan  !

Leta 1 1 = ; + 2 1 2

1 1

1

1

1

1

=  maka

= 20 40 = 20 1 1 1 = + = 20 +

2

2

1 = tak hingga

2

2

1

=

2

1

=  = 40 cm 1

1 1  maka 40 2

=

1 40

1

1

+ 20 = 40

= 40

1 jika s2

2

1

1

1 1

1

1

=  ; + =

1

2

= =

= 20 1 1 = 40 2

1

1

2

=

1  ; maka 2 40

= 40

( 4 0 ) = 60 1 1 = 60 120

= 120

120 = 80 60 1 40 2 = 1 = 40 = 80 20 2 Jadi, letak titik fokus benda (F) adalah 120 cm di depan lensa positif. = = 80 =

1.

2

1

= 40

Dua celah yang berjarak 1 mm, disinari cahaya merah dengan panjang gelombang 6,5x10-7 m. Garis gelap terang dapat diamati pada layar yang berjarak 1 m dari celah. Hitunglah jarak antara gelap ketiga dan terang pusat, serta jarak antara terang kedua dengan garis terang keempat! Penyelesaian: Diketahui: d = 1 mm = 10-3 m -7 m l=1m Ditanya: a. p

λ = 6,5 x 10 b. Δp  − −  λ  − −  λ  −  

Pembahasan: a. Jarak antara gelap ketiga dengan terang pusat 1 = 2 1 l 1 6,5×10 7 × 1 = = 3 2 2 10 3

−−



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  ∆ ∆λ −    ∆λ − ∆∆  −  −   − sebuah kisi difraksi. Interferensi maksimum terjadi dengan membentuk sudut 30˚. λθ==945030˚ Å = 9450 x 10   λ λ = panj a ng gel o mbang      − − −  −   −  −    dan gar i s or d e 1 adal a h 12˚ sin 12˚ = 0,208. Berapa panjang gelombang cahaya tersebut?     −  −   b.

= 1,6 Jarak antara terang kedua dan terang keempat =

=

l

4

=

= 1,3 × 10

2.

3

2 6,5 × 10 10 3 = 1,3

7

1

=13×10

4

Seberkas sinar mempunyai panjang gelombang 9450 Å ditujukan tegak lurus pada

Berapa banyak goresan goresan pada kisi tersebut setiap cm? Penyelesaian: Diketahui: 1 Å = 10-10 m -10 m Ditanya: d Pembahasan: sin = Keterangan: Keterangan: d = jarak antara goresan atau celah n = orde sin30 s in30 = 1 9450 × 10 10 0,5 0,5 = 1 9450 × 10 10 9450 × 10 10 = 0,5 = 18900 × 10 10 = 18 9 0 0 × 1 0 8 Banyak goresan atau celah tiap1 cm: 1 = 52 5291 91 / 189900 × 10 8 18

3.

Sebuah kisi yang mempunyai 2000 garis setiap cm, digunakan untuk menentukan panjang gelombang cahaya. Sudut antara garis pusat

Penyelesaian: Diketahui: 1 = 2000 = 12°

=5×10

4

=5×10

6

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  − λ   λ λλ == 1,1040004 x 10Å sin = 5×10 4

4.

0,208 0,208 = 1 -6 m

Sebuah celah ganda disinari dengan cahaya yang mempunyai panjang gelombang 560 nm. Sebuah layar diletakkan 1 m dari celah. Jika jarakantara kedua celah 0,5 mm, maka berapa jarak dua pita terang yang y ang berdekatan? Penyelesaian: Diketahui: 1 nm = 10-9 m -9 m l=1m d = 0,5 mm = 0,5 x 10 -3 m Ditanya: y Pembahasan:

λ = 560 nm = 560 x 10

  λ    −−   − −    − −  −     −−    −  sin =

1

=

1

sin =

0,5 × 10

3

0,5 × 10 3 0,5 × 10 3 560×10 5 60×10 = 0,5 × 10 = 1120 × 10 = 1120 × 10 = 1,12 ,120

5.

1

9

= 1 560 × 10

9

= 1 1 560×10 5 60×10 = 560×10 9 2

9

2

2

3

4 3

Cahaya monokromatik melewati dua celah sempit yang sejajar. Jarak antara kedua celah adalah 0,6 mm. Jarak antara layar dengan kedua celah adalah 60 cm. Pola interferensi yang terjadi pada layar adalah berupa garis terang dan gelap yang dipisahkan oleh jarak yang sama. Jika jarak dua garis g aris terang berdekatan berdekatan adalah 0,2 mm. Tentukan panjang gelombang cahaya yang digunakan! Pembahasan: Diketahui: d = 0,6 mm = 0,0006 m = 6 x 10-4 m y = 0,2 mm = 0,0002 m = 2 x 10 -4 m l = 60 cm = 600 mm = 0,6 m Ditanya: Penyelesaian:

λ

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 λ λ    −  λ = 19,8 x 10

−

= sin 1 = 6 × 1 0 4  m 3,3×10 4 -8 m = 1,98 x 10 -7 m = 198 nm

adalah: y = 10 sin π 2t –

1. Persamaan gelombang transversal yang merambat di tali 0,0x) dimana x dan y dinyatakan dalam cm dan t dalam detik. Tentukan: a) Amplitudo, frekuensi, kecepatanfasa, kecepatanfasa, dan panjang gelombang b) Kecepatan Kecepatan maksimum transversal suatu partikel di dalam tali Jawab: Persamaan soal di atas dapat ditulis x t  x - 200 = A si sin 2  (T ) a) Amplitudo A = 10 cm Periode T = 1 detik Panjang -1 Kecepatan x b) - 200 ). Ambil partikel pada x = 0 dan GHS yang dialami partikel tersebut adalah:

y = 10 sin 2π t π − λ gelfoambang λ = 200 cm s a c = λf = 200 cm. s Dari persamaan y = 10 sin 2π t y = 10 sin 2πt = 2π. 10 cos 2πt = 20π cos 2πt dan maks = 20π cm

Turunkan y terhadap waktu t menghasilkan kecepatan partikel di dalam tali (dalam arah tegak lurus) dy vy = dt   vy .s-1

2. Gelombang bunyi dengan frekuensi 500 Hz mempunyai kecepatan 350 ms -1 dan amplitudonya 0,3 mm. a) Tuliskan persamaan gelombangnya

fk==500ω/cHz,= 1000π/350 ω = 2πf = 1000π r a d. s m Jadi y = 0,3 sin 1000 πt π π = 0, 3 s i n 1000 πt ,  m m seπjar′ak x’π, y’ = 0,3 sin 1000πt π x’ x’ = 0,117gelmombang λ = dan x’ = λ = λ/6 untuk fasa tertinggal -1

-1

x

-350 ) (mm) b) Seberapa jauh dari titik asal, suatu titik lain dengan fasa tertinggal 60 0? Catatan: SEMUA SUDUT HARUS HARUS DINYATAKAN DALAM RADIAN 2 600 = 60 X 360  = 3  radian Di titik asal( x = 0), y0 1000 Di titik - 350 Jadi,

1000 350

Panjang

x =

3

c f 

350

= 500 = 0,7 0,7 m

0,117 0,7

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Di titik x = λ/4 = 0,7/4π= 0,175 m, maka Di titik x = 0, maka y = 0 y = 0,3 sin (

1000 350

π

.0,175)

y = 0,3 sin 2 = 0,3 0,3 mm mm 3. Gelombang transversal merambat dari kiri ke kanan dengan panjang gelombang 10 m dan kecepatan 20 ms-1. Amplitudo gelombang adalah 0,25m, dan ujung gelombang pada titik x = 0, berada dalam keadaan seimbang y = 0 pada saat t = 0. Tuliskan persamaan persamaan gelombang, anggap gelombangnya sinusoidal. Persamaan yang dicari secara umum berbentuk  kx Dan pada x = 0 dan t = 0, maka y = 0, sehingga 0 = A sin (-

y = A sin ωt – – φ φ φ = 0, π, 2 π, 3 π, …………. Ambi l φ = 0 pi l i h an pal i n g maka y = A s i n  ω t – ω = 2πλ π π π −− π λπ π π − dengan x,y dal= Aamsinmetωetr–dan–t πdal=am Adetsiink.Ambi l kemudi a n φ = π, maka ω t – 0,25 sin 4πt –0,– 0,2πx at u dawai y = 3 s i n π – adalah …. 

 sederhana  kx) dengan A = 0,25 m, c 2 .20 2 2 = 10 = 4  rad.  rad.ss 1  da  dan k =  ; k = 10 = 5 m 1 Jadi, y = 0,25 sin (4 t  0.2 x) merupakan persamaan gelombang yang dicari, Y

 kx

-

 kx)

=-

4. Persamaan gelombang transversal yang merambat pada su (120 t   0,4 x). Jika x dan y dalam cm dan t dalam detik, maka panjang gelombangnya A.1cm B.2cm C.3cm D.4cm E.5cm Pembahasan : Tiga bentuk  yang merambat pada arah sumbu x positif (ke kanan).

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gelombang, v = cepat rambat gelombang Diketahui :

 : A = 3 cm  : t / T = t f = 60 t f = 60 hertz :

Ditanya : Panjang gelombang ? Jawab : Panjang gelombang = 5 cm Jawaban yang benar adalah E. 5. Gelombang berjalan merambat pada tali ujung tetap dilukiskan seperti pada diagram di bawah ini :

tA.ray=nsv0,er4ssailnyangπ 3mert – ambat ke kanan adalah…

Jika jarak AB = 3 meter ditempuh selama 1 sekon, maka persamaan gelombang gelombang  x) m

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Ditanya : Persamaan Persamaan gelombang ? Jawab :

1. Jelaskan pengertian difraksi dan gambarkan pola difraksi dari berkas sinar yang melewati celah sempit berbentuk lingkaran. Difraksi merupakan perambatan atau pembelokan arah rambat cahaya yang menimbulkan pola terang dan gelap dimana intensitas pola terang semakin jauh maka semakin kecil pula intensitasnya.

2. Bagaimana nilai N maksimal dan nilai N minimal yang terjadi pada difraksi celah

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3. Bagaimana syarat terjadinya garis gelap dan garis terang pada difraksi celah tunggal?

    …          …

Syarat terjadinya garis gelap ke-m adalah: sin = ; = 1,2, 1,2,3, 3, Untuk sudut  yang kecil, berlaku : =

Syarat terjadinya garis terang ke-m adalah: 1 sin = + ; = 0,1, 0,1,2, 2, 2

4. Apabila masing-masing celah ganda memiliki dimensi lebar b dan panjang l (b