analisis dataprisma
DESCRIPTION
praktikumfisdas2TRANSCRIPT
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
a) Sudut Deviasi Minimum
δ k=T k- T 0
δ k=47,98333°−0,00000 °
δ k=47,98333°
b) Analisis Ketidakpastian
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°
c) Kesalahan Relatif
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
a) Sudut Deviasi Minimum
δ b=T b-T 0
δ b=48,23333 °−0,00000 °
δ b=48,23333 °
b) Analisis Ketidakpastian
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°
c) Kesalahan Relatif
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
b) Analisis Ketidakpastian
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
b) Analisis Ketidakpastian
∆ 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
c) Kesalahan Relatif
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
b) Analisis Ketidakpastian
∆ 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
c) Kesalahan Relatif
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|