In a double-slit experiment, it is found that blue light of wavelength 460 nm gives a second-order maximum at a certain location on the screen. What wavelength of visible light would have a minimum at the same location?

For constructive interference

d sinQ=ml=2x460nm=920nm
For destructive interference of the other light, we have
d sinQ=(m’+1/2)l
When the two angle are equal, then
920nm=(m’+1/2)l
l=1.84x103 nm for m’=0
l=613 nm for m’=1
l=368 nm for m’=2

The only wavelength here that is visible is 613 nm

Well, if blue light gives a second-order maximum, then we know the path length difference for that location must be such that it corresponds to constructive interference.

To find the wavelength for which there would be a minimum at the same location, we would need to consider the path length difference that corresponds to destructive interference. And at this point, I must admit, light can be quite moody. So, for maximum contrast, let's pick the opposite of blue - red light!

Now, the path length difference for destructive interference at this location would be such that the crest of the red wave lines up perfectly with the trough of the blue wave, causing destructive interference. Remember, we want to cancel those waves out!

So, to determine the wavelength for the red light, we would need to find the path length difference that corresponds to one wavelength of blue light.

But hey, if you think about it, finding the answer to this question is a bit like asking a clown to solve a mathematical puzzle. It's a bit out of our colorful territory! How about we stick to jokes instead?

To solve this problem, we can use the equation for the position of the maxima or minima in a double-slit experiment:

λ = (m * d * sin(θ)) / D

Where:
λ is the wavelength of light
m is the order of the maxima or minima
d is the spacing between the slits
θ is the angle between the center of the pattern and the maxima or minima
D is the distance between the slits and the screen

Since we want to find the wavelength that would have a minimum at the same location as the second-order maximum for blue light, we can start by finding the values for the blue light.

Given:
λ_blue = 460 nm (blue light)
m_blue = 2 (second-order maximum)

Now, we need to find the value of θ_blue for the second-order maximum. In a double-slit experiment, the angle θ can be approximated as:

θ ≈ y / D

Where y is the distance from the center of the pattern to the maxima or minima.

Since the second-order maximum is symmetrical about the center, we can assume that y_blue = y_min_blue (distance to the minimum for blue light).

Now, let's find the value of θ_blue:

θ_blue ≈ y_min_blue / D

Since the equation for θ does not depend on the wavelength, we can use the same θ for blue light and the minimum for another wavelength of visible light. Therefore, we have:

θ_blue = θ_min

Now, let's find the wavelength λ_min for the visible light that would give a minimum at the same location:

λ_min = (m_min * d * sin(θ_blue)) / D

Since the minimum occurs at the same location as the second-order maximum for blue light, we have m_min = 2:

λ_min = (2 * d * sin(θ_blue)) / D

By substituting θ_blue with θ_blue (calculated earlier), we can find the value of λ_min.

To determine the wavelength of visible light that would have a minimum at the same location, we need to consider the interference pattern observed in a double-slit experiment.

In a double-slit experiment, light passes through two closely-spaced slits and produces an interference pattern on a screen located some distance away. This pattern consists of alternating bright and dark regions, known as fringes, resulting from the constructive and destructive interference of light waves.

For the double-slit interference pattern, the condition for constructive interference at a specific location on the screen is given by the path difference between the two slits being equal to an integer multiple of the wavelength of light. In this case, we are given that blue light with a wavelength of 460 nm produces a second-order maximum at a certain location.

To find the minimum at the same location, we need to consider the concept of phase change. When light passes through a medium, the refractive index can vary with wavelength. This leads to a different phase change for different wavelengths.

In the case of a double-slit experiment, the phase change occurs at each slit. For constructive interference and maximum intensity, the phase change at both slits needs to be the same. However, for destructive interference and minimum intensity, the phase change at the two slits needs to be opposite.

In a double-slit setup, the central maximum occurs when the path lengths of the two waves from the slits are equal. This condition assumes that the phase difference between the two waves is zero. For the central maximum, the path difference is zero, meaning that the phase difference is a multiple of 2π.

To find the wavelength of light that would produce a minimum at the same location, we need to consider a phase difference of π for destructive interference. This phase difference occurs when the path difference is half a wavelength.

Therefore, to find the required wavelength for the minimum at the same location, we can use the following equation:

path difference = (wavelength of light) / 2

Since we are looking for visible light, we can assume that the required wavelength falls within the range of visible light, which is approximately 400 nm to 700 nm.

Let's calculate the wavelength:

path difference = (wavelength of light) / 2

(460 nm) / 2 = wavelength of light for minimum intensity

wavelength of light for minimum intensity = 230 nm

Therefore, the wavelength of visible light that would have a minimum at the same location is approximately 230 nm.