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2 years ago

Why are choir concerts better in performance halls than in gymnasiums?

1 answer:

3 0

Constructive interference in performance halls and the elimination of echoes are reasons why choir concerts better in performance halls than in gymnasiums.

<h3>What is constructive interference?</h3>

Constructive interference is a phenomenon which occurs when two waves travelling in same direction and which are in phase add up together to produce a wave of greater amplitude.

Constructive interference occurs in performance halls while destructive interference occurs in gymnasiums.

Also in performance halls, echoes are minimized due to the padded walls and the curtains while echoes which disturb choir concerts occur in gymnasiums.

Therefore, choir concerts better in performance halls than in gymnasiums because of constructive interference in performance halls and the elimination of echoes.

Learn more about constructive interference and echoes at: https://brainly.in/question/2378717

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Answer:

(a) d = 1960nm

(b) The slit should be decreased.

(c) Δd = 360nm.

Explanation:

The double-slit interference is given by the following equation:

$d sin(\theta) = m \lambda$ (1)

<em>where d: is the distance between slits, Θ: is the angle between the path of the light and the screen, m: is the order of the interference and λ: is the wavelength of the light. </em>

(a) To determine the least wavelength in the visible range in the third-order we need first to find the distance between slits, using equation (1) for a fourth-order:

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 490nm}{sin(90)} = 1960nm$

Now, we can find the least wavelength in the visible range in the third-order:

$\lambda = \frac{d sin(\theta)}{m} = \frac{1960nm sin(90)}{3} = 653nm$

So, the least wavelength in the visible range (400nm - 700nm) in the third-order is 653nm.

(b) To eliminate all of the visible light in the fourth-order maximum <u>means that the wavelength must be smaller than 400nm</u>, and hence the slit separation should be decreased <u>since they are proportional to each other</u> (see equation (1)).

(c) The distance between slits needed to eliminate all of the visible light in the fourth-order maximum, with λ = 400 nm as limit value, is:

$d = \frac{m \lambda}{sin(\theta)} = \frac{4 \cdot 400nm}{sin(90)} = 1600nm$

Therefore the least change in separation needed is equal to the initial distance calculated for 490nm and the final distance calculated for 400nm:

$\Delta d = d_{i} - d_{f} = 1960nm - 1600nm = 360nm$

I hope it helps you!

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