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Effectus [21]
3 years ago
14

Sunlight is used in a double-slit interference experiment. The fourth-order maximum for a wavelength of 490 nm occurs at an angl

e of θ = 90°. Thus, it is on the verge of being eliminated from the pattern because θ cannot exceed 90° in Eq. 35-14. (a) What least wavelength in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all of the visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change in separation is needed?
Physics
1 answer:
Black_prince [1.1K]3 years ago
3 0

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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