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nadya68 [22]
2 years ago
5

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

m thickness for which destructive interference of reflected light is observed is 278 nm. What is the vacuum wavelength of the light in nm?
Physics
1 answer:
Anika [276]2 years ago
4 0

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

2t = 2\lambda_{film}

t = \lambda_{film}

The index of refraction of soap is given, then

\lambda_{film} = \frac{\lambda_{vacuum}}{n}

Combining the results of all steps we get

t = \frac{\lambda_{vacuum}}{n}

Rearranging, we find

\lambda_{vacuum} = tn

\lambda_{vacuum} = (278)(1.33)

\lambda_{vacuum} = 369.74nm

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a) p_i=1.568\hat{i}+0.752 \hat{j}

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Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

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Initial momentum:

p_i=m_1.v_1+m_2.v_2

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p_i=1.568\hat{i}+0.752 \hat{j}

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p_i=\sqrt{1.568^2+0.752 ^2}

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m_f.v_f=1.739

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Direction of final velocity will be in the direction of momentum:

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\theta=25.62^{\circ}

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v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

6 0
3 years ago
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