Red light with wavelength 700 nm is passed through a...

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe ($m$ = 32) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As
a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe ($m$ = 32) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

Key Concepts

Parameter Independence – Slit Spacing Cancellation

Although the interference condition involves the slit spacing d, when comparing the positions of fringes for two different wavelengths the spacing appears on both sides of the relevant equations and cancels out. This means that determining whether a given fringe for one wavelength coincides or does not coincide with a fringe from the other wavelength can be done solely in terms of the wavelengths and fringe order indices. Consequently, the specific value of the slit spacing is not necessary to answer the question about possible wavelengths of the second light.

Rational Ratios and Purity of a Fringe

If the wavelengths are in a rational ratio, then many of the bright fringes for both lights can nearly coincide, creating a mixed color appearance. However, for a particular fringe of one light (for example, the red light’s m = 32 fringe) to appear pure, the corresponding value (32×?_red divided by ?_other) must not be an integer. In other words, the other wavelength must be such that its interference maximum does not occur exactly at that fringe, resulting in a pure single color there.

Fringe Coincidence and Wavelength Ratio

When two wavelengths are present in a double?slit experiment, the positions of their bright fringes are given by their respective interference conditions. For the fringes of the two colors to overlap, the condition m??? = m??? must be satisfied for some integers m? and m?. The ratio of the wavelengths therefore determines how closely the two sets of fringes coincide. Slight discrepancies in this ratio can lead to most fringes overlapping while leaving isolated fringes where only one wavelength contributes.

Double?Slit Interference Condition

In a double?slit experiment, bright fringes on the screen occur at angles where the path difference between light coming from the two slits is an integer multiple of the wavelength. This is expressed by the principle d sin? = m?, where d is the slit spacing, ? is the angle relative to the central maximum, m is an integer (the order number), and ? is the wavelength of the light. This formula underpins the formation of the interference pattern for any monochromatic light source.

Transcript

00:01 Okay, so here, light of wavelength lambda, let me write that here, the light of wavelength lambda r is equal to 700 nanometers, is blast passed through us to set apart us.

00:14 Now, at the same time, another monochromatic light is passed.

00:17 Let us consider that to be lambda, the wavelength of light.

00:20 Now, it was seeing that the center of the third bright fringe appears pure rate.

00:28 So that means that the that position at that particular position the real light the lambda r went on to have constructive interference whereas lambda went on to have destructive interference let me write that here at that particular point where it was red lambda r underwent constructive interference whereas lambda underwent destructive interference that's why we can see pure right so for that position if that we consider that to be y then why should be equal to three times lambda r d by d where capital d is the distance between let me write it here so distance between slit and screen or rise small d is the slit spacing right so this is for the red light and at the same point why there is destructive interference due to lambda so for destructive difference we write m called to half m plus half of lambda by d all right so from these two equations what we can figure out is that from these two equations you can write m plus half of lambda should be equal to three times lambda for rate right so lambda equals to three upon m plus half of lambda r right so let us turn by term the values for m m equal to zero gives lambda equals to 4200 nanembras well this is not in visual reason we need we need wavelengths in visual region.

03:01 So let's move on.

03:03 For m equal to 1, lambda becomes 1 ,400 nanometers.

03:08 Okay, it is decreasing, right? so am equal to 2...

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