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andrew-mc [135]
2 years ago
9

Find the distance in nm between two slits that produces the first minimum for 410-nm violet light at an angle of 14.5°.

Physics
2 answers:
docker41 [41]2 years ago
8 0

Answer:

Explanation:

wavelength, λ = 410 nm

Angle, θ = 14.5°

The formula used for the distance between the two slits for minima is given by

dSin\theta =\left (m+\frac{1}{2}  \right )\lambda

for first order minima, m = 0

d Sin 14.5 = 0.5 x 410

d = 818.76 nm

Galina-37 [17]2 years ago
5 0

Answer:

820 nm

Explanation:

We are given that

Wavelength=\lambda=410 nm

\lambda=410\times 10^{-9} m

1nm=10^{-9} m

\theta=14.5^{\circ}

For first minimum therefore

m=0

We know that for destructive interference

(m+\frac{1}{2})\lambda=dsin\theta

Substitute the values

(0+\frac{1}{2})\times 410\times 10^{-9}=dsin 14.5

d=\frac{410\times 10^{-9}}{2\times sin 14.5}

d=820\times 10^{-9} m=820 nm

Hence, the distance between two slits that produces the first minimum=820 nm

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The given parameters:

  • <em>Length of the string, L = 100 cm</em>

<em />

The wavelengths of the constituent travelling waves is calculated as follows;

L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}

for first mode: n = 1

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for second mode: n = 2

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\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm

For fourth mode: n = 4

\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50  \ cm

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The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

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