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Lynna [10]
3 years ago
8

A single slit forms a diffraction pattern, with the first minimum at an angle of 40.0° from central maximum, when monochromatic

light of 630-nm wavelength is used. The same slit, illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum. What is the wavelength of this new light?
Physics
1 answer:
Margarita [4]3 years ago
3 0

Answer:

\lambda = 424.4 nm

Explanation:

As we know by the formula of diffraction

a sin\theta = N\lambda

so we have

a = slit size

\theta = angular position of Nth minimum

so we will have

for first minimum of 630 nm light

a sin40 = 1(630 \times 10^{-9})

a = 9.8 \times 10^{-7} m

Now for another wavelength second minimum is at 60 degree angle

a sin60 = 2 \lambda

(9.8 \times 10^{-7}) sin60 = 2 \lambda

\lambda = 424.4 nm

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Answer:

The initial acceleration of the 59g particle is 1062.7\frac{m}{s^{2}}

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

F=ma (1)

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F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}

with q1 and q2 the charge of the particles, r the distance between them and k the constant k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}. So:

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Using that value on (1) and solving for a

a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}

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