## User: neeta.vanage

neeta.vanage •

**60**- Reputation:
**60**- Status:
- Trusted
- Location:
- Last seen:
- 3 months ago
- Joined:
- 5 months, 1 week ago
- Email:
- n***********@gmail.com

#### Academic profile

None

None

None

None

#### Posts by neeta.vanage

<prev
• 261 results •
page 1 of 27 •
next >

0

votes

1

answer

211

views

1

answers

... N = 4000 line/cm, λ = 5000 x10$^{-8}$, n= 3
$
\frac{dθ}{dλ}= \frac{n}{(a+b)cosθ} \\[3ex]
(a+b) sinθ = nλ \\[2ex]
sinθ = \frac{nλ}{(a+b)} \\[2ex]
sinθ = N nλ \,\,\,\,\,\ [\because (a+b) = \frac{1}{N}] \\[2ex]
sinθ = 4000 \times 3 \times 5000 \times 10^{-8} = 0.6 \\[2ex]
θ= 36.86
$
cosθ = ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

191

views

1

answers

... RP = $\frac{λ}{dλ}$ = nN
For wavelengths 5140.34 & 5140.85 Å……..
Mean wavelength λ is $\frac{5140.34 + 5140.85}{2} = 5140.595 \ A° $
Smallest difference between them is 5140.85 – 5140.34 = 0.51 A°
First order, n=1
$ N= \frac{1}{n} \times \frac{λ}{dλ} = \frac{1}{1} \times \frac{5140.59 ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

118

views

1

answers

... Width = 0.5 cm, N = 5000 lines /cm
Therefore total number of lines on grating is $\frac{5000}{0.5} = 10000 $
Mean wavelength λ is $\frac{ 5890.2 + 5896.4}{2} = 5893.3 \ A° $
Smallest difference between them is 5896.4 – 5890.2 =6.2 A°
$ RP = \frac{λ}{dλ} = \frac{5893.3}{6.2} = 950.5 $
Als ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

136

views

1

answers

... N = 10000 lines/cm, Width W = 5 cm, λ = 6000 Å, n = 2, RP = ?, dλ = ?
Total number of lines on grating are N = 10000x 5 = 50000
RP = nN
RP = 2x 50000
RP = 100000
$\frac{λ}{dλ}$ = RP
100000 = $\frac{6000 \times 10^{-8}}{dλ}$
dλ = 0.06 A°
...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**1

vote

1

answer

123

views

1

answers

... RP = $\frac{\lambda}{d\lambda}$ = nN
$λ = \frac{5890+5896}{2} = 5893 A° $
dλ = 5896-5890 = 6 A°
$\frac{λ}{dλ}$ = 982
982 = 1 x N
Therefore N=982 lines /cm
For grating of width 2 cm N = 982 x2 = 1964
...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

118

views

1

answers

... Width W = 3 cm, N = 7000 lines/cm, λ = 5000 A°
grating element $ (a+b) = \frac{1}{N} = \frac{1}{7000} $
RP is maximum when n is maximum
N is maximum when sin θ =1
(a+b) sin θ = n λ
$ n_{max} = \frac{1}{Nλ} = \frac{1}{ 7000 \times 5000 \times 10^{-8}} = 2.8 $
n cannot be 3 as s ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**1

vote

1

answer

149

views

1

answers

...
$ D_{(n+p)}^2 - D_n^2 = \frac{4pλR}{μ} $
For air film, μ = 1
For 12$^{th}$ and 4$^{th}$ dark rings:
$ D_{12}^2 - D_4^2 = 4 \times 8 \times λ \times R $ ………………………………..(1)
For 20$^{th}$ and 4$^{th}$ dark rings:
$ D_{20}^2 - D_4^2 = 4 \times 16 \times λ \times R $ ……………………………….(2)
Dividin ...

written 4 months ago by
neeta.vanage •

**60**• updated 11 weeks ago by Yashbeer ♦♦**80**0

votes

1

answer

213

views

1

answers

... RP = $\frac{λ}{dλ}$ = nN
dλ = $\frac{λ}{nN}$
Velocity = frequency x wavelength
C= νλ
$ ν = \frac{c}{λ}$……………………………(1)
Differentiate the equation 1
$ dν = -c \frac{dλ}{λ^2} $………………………………(2)
Substituting equation 1 in equation 2,
$ dν = -c \frac{λ}{nN λ^2} $
Therefore, $ dν = -\ ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

128

views

1

answers

... (a+b) = $\frac{2.54}{15000}$, wavelength λ$_1$ = 4000 AU
$
(a+b) sinθ = n λ_1 \\[2ex]
sinθ= \frac{n λ_1}{(a+b)} \\[2ex]
θ= sin^{-1} (\frac{n λ_1}{a+b}) \\[2ex]
θ= sin^{-1} (\frac{n \times 4000 \times 10^{-8}}{2.54/15000}) \\[2ex]
θ= 0.236 \times n \\[2ex]
$
Substitute value of n as 1,2,3
An ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Manan Bothra •**0**0

votes

1

answer

136

views

1

answers

... $$ $$
$B =10^-4 Wb/m^2 \\
l = 3 cm, \\
d = 20 cm. \\
V_A = 800 V,\\
Deflection =?, \\ $
$\text{Charge of an electron }1.6 *10^{-19} \\ $
$ \text{Mass of an electron } 9.1*10^{-31} \\ $
$ Deflection Y = DlB √(e/(2mVA )) \\ $
$ Y = 20 * 10^-2 * 3 * 10^-2 *1 x 10^-4 √(( 1.6 *10^{-19})/(2 * 9.1 ...

written 4 months ago by
neeta.vanage •

**60**• updated 4 months ago by Saurabh Singh •**0**#### Latest awards to neeta.vanage

Centurion
5 months ago,
created 100 posts.

Rising Star
5 months ago,
created 50 posts within first three months of joining.

Popular Question
5 months ago,
created a question with more than 1,000 views.
For What is the Rayleigh`s criteria of resolution? Define resolving power and derive the expression for the resolving power of grating.

Popular Question
5 months ago,
created a question with more than 1,000 views.
For Obtain the condition for maxima and minima due to interference in a wedge shape film observed in reflected light. How is the interference pattern in wedge shaped film?

Great Question
5 months ago,
created a question with more than 5,000 views.
For Obtain the condition for maxima and minima due to interference in a wedge shape film observed in reflected light. How is the interference pattern in wedge shaped film?

Site

- Publications
- Advertise
- RSS
- Stats

Use of this site constitutes acceptance of our User
Agreement
and Privacy
Policy.