## User: neeta.vanage

neeta.vanage60
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#### Posts by neeta.vanage

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... 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 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0
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... 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 $N= \frac{1}{n} \times \frac{λ}{dλ} = \frac{1}{1} \times \frac{5140.59 ... written 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0 1 answer 212 views 1 answers ... Width = 0.5 cm, N = 5000 lines /cm Therefore total number of lines on grating is$\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 $Als ... written 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0 1 answer 230 views 1 answers ... N = 10000 lines/cm, Width W = 5 cm, λ = 6000 Å, n = 2, RP = ?, dλ = ? Total number of lines on grating are N = 10000x 5 = 50000 RP = nN RP = 2x 50000 RP = 100000$\frac{λ}{dλ}$= RP 100000 =$\frac{6000 \times 10^{-8}}{dλ}$dλ = 0.06 A° ... written 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0 1 answer 227 views 1 answers ... RP =$\frac{\lambda}{d\lambda}$= nN$λ = \frac{5890+5896}{2} = 5893 A° $dλ = 5896-5890 = 6 A°$\frac{λ}{dλ}$= 982 982 = 1 x N Therefore N=982 lines /cm For grating of width 2 cm N = 982 x2 = 1964 ... written 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0 1 answer 195 views 1 answers ... Width W = 3 cm, N = 7000 lines/cm, λ = 5000 A° grating element$ (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 $n cannot be 3 as s ... written 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0 1 answer 884 views 1 answers ...$ 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 $……………………………….(2) Dividin ... written 9 months ago by neeta.vanage60 • updated 8 months ago by Yashbeer ♦♦ 130 1 answer 469 views 1 answers ... RP =$\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 9 months ago by neeta.vanage60 • updated 9 months ago by Manan Bothra0
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... (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 ...
...  $B =10^-4 Wb/m^2 \\ l = 3 cm, \\ d = 20 cm. \\ V_A = 800 V,\\ Deflection =?, \\$ $\text{Charge of an electron }1.6 *10^{-19} \\$ $\text{Mass of an electron } 9.1*10^{-31} \\$ $Deflection Y = DlB √(e/(2mVA )) \\$ \$ Y = 20 * 10^-2 * 3 * 10^-2 *1 x 10^-4 √(( 1.6 *10^{-19})/(2 * 9.1 ...