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User: neeta.vanage

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neeta.vanage60
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Posts by neeta.vanage

<prev • 261 results • page 1 of 27 • next >
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Answer: A: A diffraction grating have 4000 lines per cm is illuminated normally by light of
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: Will it resolve the lines 8037.2 & 8037.5 $\mathring{A}$ in the $2^{nd}$ order?
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: Would you expect these lines to be distinguished by the grating in the $1^{st}$
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: A grating has 10000 lines/cm & its total width is 5 cm. Calculate for $\lambda$
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: Find the minimum number of lines required in a grating of width 2 cm so that the
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: Find the maximum value of RP a grating 3 cm wide & having 7000 lines/cm can have
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: In the Newtons ring experiment, determine the diameter of the $20^{th}$ ring if
... $ 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 7 months ago by neeta.vanage60 • updated 5 months ago by Yashbeer ♦♦ 100
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Answer: A: The resolving power (RP) of a diffraction grating is given by $\lambda /d\lambda
... 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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: White light (400 nm <$\lambda$< 700 nm) is normally incident on a grating. Show
... (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 7 months ago by neeta.vanage60 • updated 7 months ago by Manan Bothra0
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Answer: A: Also find the displacement of a particle with charge twice that of an electron
... $$ $$ $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 7 months ago by neeta.vanage60 • updated 7 months ago by Saurabh Singh0

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