# Double Slit Interferance Intensity

• JJK1503
In summary, the problem involves a Helium laser with a wavelength of 588 nm shining on double-slits separated by 1.80 mm. An interference pattern is observed on a screen and the point C on the screen is at the center of the principal maximum. The point P is the point on the principal maximum where the intensity of light is half that of the intensity at C. Using the small angle approximation, the value of the angle θ is found to be approximately 0.004679 degrees.
JJK1503

## Homework Statement

A Helium laser, λ = 588 nm, shines on double-slits separated by 1.80 mm. An interference pattern is observed on a screen at a distance R from the slits. The point C on the screen is at the center of the principal maximum of the interference pattern. The point P is the point on the principal maximum at which the intensity of light is half that of the intensity at C. What is the value of the angle θ?

## Homework Equations

I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )

small angle approximation sin(θ) = tan(θ) = θ

## The Attempt at a Solution

On questions like these I typically find it easiest to pick some dummy value and solve. In this case i set I (initial) to 10.

The center of the center maxima is where intensity is the greatest. At this point θ = 0.
so,
I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )
= 10 * cos^2( (pi * d * sin(0)) / λ )
= 10 * cos^2 (0)
= 10

If I want the θ where I (final) = I(initial0 / 2. I need to find θ where I (final) = 10 / 2 = 5
so (with small angle approximation),
I (final) = I (initial) * cos^2( (pi * d * sin(θ)) / λ )
5 = 10 * cos^2( (pi * (1.8 * 10^-3) * θ) / (588 * 10^-9 )
0.5 = cos^2( (pi * (1.8 * 10^-3) * θ) / (588 * 10^-9 )
0.5 = cos^2 (9617.1204 * θ)
At this point I plug the above equation into my calculator and have it solve for θ.

According to my calculator θ = 0.03743 deg

This seems like a reasonable result; However, the computer kicks it out as incorrect.

I am NOT looking for someone to give me this answer. However, ANY help with my method is GREATLY appreciated.

Last edited:
Everything you are doing is correct. You have but one problem at the end though. You are probably typing it in your calculator wrong.

0.5 = cos^2 (9617.1204 * θ)
According to my calculator θ = 0.03743 deg

0.5 = cos^2 (9617.1204 * θ)
sqrt(0.5) = cos (9617.1204 * θ)
cos^-1(sqrt(.5))=9617.1204θ
θ=.004679 degYou asked me to not supply you the answer but All I am doing is being your calculator as you did everything else correctly ;)

Seigi Yokota said:
Everything you are doing is correct. You have but one problem at the end though. You are probably typing it in your calculator wrong.

0.5 = cos^2 (9617.1204 * θ)
According to my calculator θ = 0.03743 deg

0.5 = cos^2 (9617.1204 * θ)
sqrt(0.5) = cos (9617.1204 * θ)
cos^-1(sqrt(.5))=9617.1204θ
θ=.004679 degYou asked me to not supply you the answer but All I am doing is being your calculator as you did everything else correctly ;)

Thank you. I appreciate you taking the time to respond. I am unsure where I made my error. My initial thought is that I might have been in radian mode, but I suspect my result would have been much smaller had that been the case. Whatever the reason; it is reassuring to know that I do indeed understand the concept.

Thanks again.

Check your calculator settings. Does the computer want the angle in radians, degrees or grads ?
Can't be a coincidence that 3743/4679 = 0.800 (but how that ratio comes about is a mystery to me -- should be the inverse ).

Why don't you just do away with the angle on the calculator in the first place: you know that ##\cos{\pi\over 4}=\sqrt 2## so you get $${\pi d \sin\theta\over \lambda} = {\pi\over 4} \quad \Rightarrow \quad \theta\approx\sin\theta = {\lambda\over 4 d}$$

## 1. What is Double Slit Interference Intensity?

Double Slit Interference Intensity is a phenomenon that occurs when a coherent light source, such as a laser, passes through two parallel slits and creates an interference pattern on a screen behind the slits. This pattern is a result of the light waves from the two slits interfering with each other, resulting in areas of constructive and destructive interference.

## 2. How is Double Slit Interference Intensity measured?

Double Slit Interference Intensity is typically measured by using a photodetector to detect the intensity of the light at different points on the screen. The intensity is then plotted as a function of position, resulting in the characteristic interference pattern.

## 3. What factors affect the intensity of Double Slit Interference?

The intensity of Double Slit Interference is affected by several factors, including the wavelength of the light, the distance between the two slits, and the distance between the slits and the screen. Additionally, the intensity can also be influenced by the angle at which the light passes through the slits and any obstructions or imperfections in the slits themselves.

## 4. What is the mathematical equation for Double Slit Interference Intensity?

The mathematical equation for Double Slit Interference Intensity is I = I0cos2(πd sinθ/λ), where I0 is the maximum intensity, d is the distance between the two slits, θ is the angle at which the light passes through the slits, and λ is the wavelength of the light.

## 5. What is the practical application of understanding Double Slit Interference Intensity?

Understanding Double Slit Interference Intensity is crucial in fields such as optics, quantum mechanics, and engineering. It can also be applied in technologies such as holography, diffraction grating, and interferometers. Additionally, studying this phenomenon can lead to a deeper understanding of light and its properties, which can have a wide range of practical applications in various industries.

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