Analisis Dataprisma

Analisis Data

Kegiatan 1

Menentukan Sudut Pembias Prisma

1. Besar Sudut Pembias Prisma (α ¿

α=T1+T 2

2

α=63,83333 °+55,16667 °2

α=¿ 59,50000 °

2. Analisis Ketidakpastian

α=T1+T 2

2

dα = 12

| ∂α∂T1|dT 1 +

12

| ∂α∂T2|dT 2

dα = 12

dT 1 + 12

dT 2 ; di mana dT 1= dT 2=¿ dT

dα = dT

∆ α=∆T

∆ α=¿ 0,01667

3. Kesalahan Relatif

KR = Δαα

× 100 %

KR = 0,01667 °

59,50000° × 100 %

KR = 0,02801 % (4 AB)

4. Pelaporan Fisika

α=¿ |α ±∆α|°

α=¿ |59,50000±0,01667|°

Kegiatan 2

A. Menentukan Sudut Deviasi Minimum

1. Untuk Spektrum Warna Merah

a) Sudut Deviasi Minimum

δm=T m - T 0

δm=47,90000 °−0,00000 °

δm=47,9000 °

b) Analisis Ketidakpastian

δm=T m - T 0

∂δm=¿ |∂δm∂T m|dT m + |∂δm∂T 0|dT o

∂δm=|1d T m+1d T 0| dimana d T m=d To= d T

d δm=2dT

∆ δm=2∆T

∆ δm=2×0,01667 °

∆ δm=¿ 0,03334°

c) Kesalahan Relatif

KR=∆ δmδm

× 100%

KR= 0,03334 °47,90000 °

× 100%

KR=0,069%(4 AB)

d) Pelaporan Fisika

δm=|δm±∆δm|°δm=|47,90000±0,03334|°

2. Untuk Spektrum Warna Kuning


δ k=T k- T 0

δ k=47,98333°−0,00000 °

δ k=47,98333°


δ k=T k- T 0

∂δ k=¿ |∂δ k∂T k|dT m + |∂δ k∂T 0|dT o

∂δ k=|1d T k+1d T 0| dimana d T k=d To= d T

d δ k=2dT

∆ δk=2∆T

∆ δk=2×0,01667 °

∆ δk=¿ 0,03334°


KR=∆ δkδ k

× 100%

KR= 0,03334 °47,98333 °

× 100%

KR=0,069 %(4 AB)

d) Pelaporan Fisika

δ k=|δk ±∆δ k|°δ k=|47,98333±0,03334|°

3. Untuk Spektrum Warna Biru


δ b=T b-T 0

δ b=48,23333 °−0,00000 °

δ b=48,23333 °


δ b=T b-T 0

∂δ b=¿ |∂δ b∂T b|dT b + |∂δ b∂T 0|dT o

∂δ b=|1d T b+1d T 0| dimana d T b=d To= d T

d δ b=2dT

∆ δb=2∆T

∆ δb=2×0,01667 °

∆ δb=¿ 0,03334°


KR=∆ δ δ bkδ b

× 100%

KR= 0,03334 °48,23333 °

× 100%

KR=0,069 %(4 AB)

d) Pelaporan Fisika

δ b=|δb±∆δ b|°δ b=|48,23333±0,03334|°

B. Menentukan Indeks Bias

1. Untuk spektrum warna Merah

a) Indeks Bias

nm= sin

12(α+δm)

sin12α

nm= sin

12(59,50000+47,90000)

sin12(59,50000)

nm= 0,80590,4962

m= 1,62414


nm= sin( α

2+δm2

)

sin12α

nm=UV

∆ nm = U 'V−UV '

V 2

∆nmnm

= [[ 1

2cos (α2 +

δm2 )∆α+ 1

2cos (α2 +

δm2 )∆δm]sin

α2– sin(α2 +

δm2 ) 1

2cos (α2 )∆α ]

sin2 α2

∆nmnm

=

[[ 12

cos (α2 +δm2 )∆α+ 1

2cos (α2 +

δm2 )∆δm]sin

α2– sin(α2 +

δm2 ) 1

2cos (α2 )∆α ]

sin2 α2

×sinα2

sin( α2+δm2

)

∆nmnm

=

12

sinα2

cos(α2 +δm2 )∆α+1

2sinα2

cos (α2 +δm2 )∆δm−1

2sin(α2 +

δm2 )cos

α2∆α

sinα2

sin( α2 +δm2 )

∆nmnm

=

12∆α [sin

α2

cos(α2 +δm2 )−sin (α2 +

δm2 )cos

α2 ]

sinα2

sin(α2 +δm2 )

−

12∆δm sin

α2

cos(α2 +δm2 )

sinα2

sin(α2 +δm2 )

∆nmnm

=¿ 12∆ α [cot(α2 +

δm2 )−cot

α2 ] - 12 ∆ δmcot(α2 +

δm2 )

∆nmnm

=12∆α cot (α2 +

δm2 )−1

2∆α cot

α2

- 12∆ δmcot(α2 +

δm2 )

∆nmnm

=|12

cot(α2 +δm2 )(∆α−∆ δm )−1

2∆α cot

α2| di mana ∆ α=∆T

∆nmnm

=|12

cot(α2 +δm2 )∆T−1

2∆α cot

α2|

∆ nm=|12∆T (cot(α2 +

δm2 ))−cot

α2 |nb

∆ nm=|12

0,01667 (cot (59,500002

+ 47,900002 ))−cot

59,500002 |1,62414

∆ nm=|12

0,01667 (cot53,7 )−cot 29,75|1,62414

∆ nm=|12

0,01667 (0,735−1,754 )|1,62414

∆ nm=0,01379

c ¿ Kesalahan Relatif

KR = ∆nmnm

× 100 %

KR = 0,013791,624

× 100 %

KR = 0,8491 % (4 AB)

d) Pelaporan Fisika

nm=¿ |nm±∆ nm|nm=¿ |1,62414±0,01379|

2. Untuk spektrum warna Kuning

a) Indeks Bias

nk= sin

12(59,50000+47,98333)

sin12(59,50000)

nk= 0,806350,4962

nk= 1,62507


∆ nk=|12∆T (cot(α2 +

δ k2 ))−cot

α2 |nb

∆ nk=|12

0,01667 (cot( 59,500002

+ 47,983332 ))−cot

59,500002 |1,62507

∆ nk=|12

0,01667 ( cot53,7 )−cot 23,99|1,62507

∆ nk=|12

0,01667 ( 0,735−2,247 )|1,62507

∆ nk=0,02047


KR = ∆nknk

× 100 %

KR = 0,020471,62507

× 100 %

KR = 1,259 % (3 AB)

d) Pelaporan Fisika

nk=¿ |nk±∆ nk|nk=¿ |1,6251±0,0205|

3. Untuk spektrum warna Biru

a) Indeks Bias

nb= sin

12(59,50000+48,23333)

sin12(59,50000)

nb= 0,80760,4962

nb= 1,62757


∆ nb=|12∆T (cot(α2 +

δ k2 ))−cot

α2 |nb

∆ nb=|12

0,01667 (cot (59,500002

+ 48,233332 ))−cot

59,500002 |1,62757

∆ nb=|12

0,01667 (cot53,7 )−cot 24,12|1,62757

∆ nb=|12

0,01667 (0,735−2,233 )|1,62757

∆ nb=0,02032


KR = ∆nbnb

× 100 %

KR = 0,020321,62757

× 100 %

KR = 1,248 % (3 AB)

d) Pelaporan Fisika

nb=¿ |nb±∆ nb|nb=¿ |1,6276±0,0203|

C. Menentukan Daya Dispersi

1. Daya Dispersi

Φ=nb−nmnk−1

Φ=1,6276−1,624141,6251−1

Φ=0,005535

2. Analisis Ketidakpastian

∆Φ=u' v−u v '

v2

∆Φ=(∆n¿¿b+∆nm) (nk−1 )−(n¿¿b−nm)(∆nk)

¿¿¿ ¿¿

∆Φ=∆nb (nk−1 )+∆nm (nk−1 )−∆nk (n¿¿b−nm)

(nk−1 ) (nk−1 )¿

∆ΦΦ

=¿¿

∆ΦΦ

=∆nb

(n¿¿b−nm)+∆nm

(n¿¿b−nm)−∆nk

(nk−1 )¿¿

∆Φ=∆nb

(n¿¿b−nm)+∆nm

(n¿¿b−nm)−∆nk

(nk−1 )×∅ ¿

¿

∆Φ= 0,0203(1,6276−1,62414)

+ 0,01379(1,6276−1,62414)

–0,0205

(1,6251−1 )×0,005535

∆Φ=0,05435

3. Kesalahan Relatif

KR=∆ΦΦ

× 100%

KR= 0,054350,005535

× 100%

KR=982% (2 AB)

4. Pelaporan Fisika

Φ=¿ |Φ± ∆Φ|

Φ=¿ |0,005±0,054|

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Analisis Dataprisma