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Lera25 [3.4K]
3 years ago
12

Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to t

he left slit than it is from the dot to the right slit

Physics
1 answer:
Bogdan [553]3 years ago
3 0

Complete question:

Suppose the wavelength of the light is 550 nm . How much farther is it from the dot on the screen in the center of fringe E to the left slit than it is from the dot to the right slit? Fringe C is the central maximum.

Check the image uploaded.

Answer:

The difference is 1100 nm

Explanation:

A bright fringe creates constructive interference, the wavelength is always in a multiple form.

At center fringe, the difference is (550 nm)(0) = 0

For the first maximum, the difference is (550 nm)(1) = 550 nm

For the second maximum, the difference = (550 nm)(2) = 1100 nm

Thus, for nth maxima, the difference is (550 nm)(n)

From the image uploaded, C is located on the second maximum, therefore the difference is given as  (550 nm)(2) = 1100 nm

Therefore, the dot on the screen in the center of fringe E to the left slit is 1100 nm

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T_{2}=278.80 K

Explanation:

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(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

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\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

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