# Intensity of central maximum when width of single slit is doubled

• songoku
songoku
Homework Statement
Relevant Equations
θ = λ / b

I ∝ ##A^2##

My answer is (D) but the correct answer is (B). I thought the intensity will be 4 times since the slit width is doubled and amplitude becomes twice. What is my mistake?

Thanks

Area of the slit is doubled. If pinhole, the area is enlarged 4 times as your speculation.

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songoku and SammyS
songoku said:
My answer is (D) but the correct answer is (B). I thought the intensity will be 4 times since the slit width is doubled and amplitude becomes twice. What is my mistake?

songoku
songoku said:
View attachment 341560

My answer is (D) but the correct answer is (B). I thought the intensity will be 4 times since the slit width is doubled and amplitude becomes twice. What is my mistake?
See equations 3.4.5 and 3.4.6 here: https://phys.libretexts.org/Courses/University_of_California_Davis/UCD:_Physics_9B__Waves_Sound_Optics_Thermodynamics_and_Fluids/03:_Physical_Optics/3.04:_Single-Slit_Diffraction

##I_0## = incident intensity.

Edit: This is a mistake. ##I_0## is not the incident intensity - so the rest of this post is wrong. See Post 5#. Apologies.

##x## = slit-width (note that '##a##' is used in the above link).
##\theta## = angle from principal axis.
Let ##\alpha = \frac {\pi x \sin\theta}{\lambda}##.

The intensity of slit’s diffraction pattern is given by ##I = I_0 [\frac {\sin \alpha}{\alpha}]^2##.

This can be written using the ‘sinc’ function as ##I = I_0~\text{sinc}^2 {\alpha}##.

For ##\theta = 0, \alpha= 0, \text{sinc}(\alpha) = \text{sinc}(0) = 1##. So the intensity at the central maximum (where ##\theta=0##) is equal to ##I_0##, which is independent of slit-width, ##x##.

This makes the answer to the original question ##I_0, \frac {\lambda}{2x}## which is not an option on the list.

Or have I made a mistake?

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Steve4Physics said:
##I_0## = incident intensity.

The intensity of slit’s diffraction pattern is given by ##I = I_0 [\frac {\sin \alpha}{\alpha}]^2##.

The ##I_0## in the formula is not the "incident intensity". That is, ##I_0## is not the intensity of the light that is incident on the slit. ##I_0## represents the intensity at the center of the central peak of the diffraction pattern. ##I_0## will depend on the slit width. I agree with @songoku and @vela that doubling the slit width doubles the amplitude of the light at the center of the central peak. Thus, ##I_0## will quadruple.

[Edit: This assumes that we are dealing with "Fraunhofer diffraction" so that the formula ##I = I_0 [\frac {\sin \alpha}{\alpha}]^2## is applicable. Otherwise, it gets much more complicated.]

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songoku and Steve4Physics
Steve4Physics said:
-------------------
It is important to understand that this expression compares the amplitude at various angles to the amplitude on the center line, equal (or approximately equal) distances from the slit. It does not provide a comparison of the amplitude of the light wave after passing through the slit to the amplitude of the plane wave before it enters the slit.
-------------------

I wonder why the author do not or cannot use intensity of incident light ?

In seciton 61 of Classical Theory of Fields by Landau and Lifshitz

where I supose LHS of the last formula udI is a typo which should be dI. RHS is different by factor ##ak/\pi## from the web author's, where a is slit width and k is wavenumber of light. The intensity is proportional to slit width. Multiplication factor is, as an examle,
$$\frac{I_{max}}{I_{incident}}=\frac{2a}{\lambda}=\frac{0.05mm}{500nm}=100$$

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songoku
anuttarasammyak said:
In seciton 61 of Classical Theory of Fields by Landau and Lifshitz... RHS is different by factor ##ak/\pi## from the web author's, where a is slit width and k is wavenumber of light. The intensity is proportional to slit width.
Landau and Lifshitz's formula for an infinitely long slit: $$dI = \frac {I_0}{\pi a k} \frac {\sin^2 ka\theta}{\theta^2} d \theta$$
The LHS of the equation, ##dI##, represents the "intensity of the diffracted light in angular range ##d \theta##". I believe this is to be interpreted as the energy per unit time that is diffracted into the angular range ##d\theta## by a unit length of the infinitely long slit. So, ##dI## has the dimensions of power per unit length of the slit.

Landau and Lifshitz (LL) define the symbol ##I_0## on the RHS as the "total intensity of the light incident on the slit". By dimensional analysis of the equation, you can check that ##I_0## has the same dimensions as ##dI##. So, I think ##I_0## must represent the total power incident on a unit length of the slit. If we let ##\mathscr{I}_0## denote the usual meaning of incident intensity as power per unit area incident on the slit, then ##I_0## in the LL equation is ##I_0 =\mathscr{I}_0 \, 2a##, where ##2a## is the width of the slit.

Thus, if we express the LL formula in terms of the incident intensity ##\mathscr{I}_0##, it reads $$dI = \frac {\mathscr{I}_0 \, 2a}{\pi a k} \frac {\sin^2 ka\theta}{\theta^2} d \theta =\frac {(2 a)^2 k \mathscr{I}_0 }{2\pi} \left( \frac {\sin ka\theta}{ka\theta}\right)^2 d \theta$$.
For ##\theta \rightarrow 0##, this gives $$dI = \frac {(2 a)^2 k}{2\pi} \mathscr{I}_0 \,d\theta$$ So, doubling the slit width ##2a## for fixed incident intensity ##\mathscr{I}_0##, will quadruple the intensity at the center of the central maximum. This is in agreement with other textbooks.

I could be wrong in my interpretation of the symbols in the LL equation. So, I welcome corrections.

songoku
TSny said:
This is in agreement with other textbooks.
Thanks. I have no other reference. Could you show me textbooks, free and online if possible.

TSny said:
andau and Lifshitz's formula for an infinitely long slit:
Thanks.
$$dI = \frac {I_0}{\pi a k} \frac {\sin^2 ka\theta}{\theta^2} d \theta$$
$$\frac{dI}{I_0} = \frac {a k} {\pi}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d \theta=\frac {2a} {\lambda}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d \theta=\frac {1} {\pi}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d (ka\theta)$$ This is an equation with no physical dimension. I wonder how your ##I_0 =\mathscr{I}_0 \, 2a## plays a role here. Say ak >> 1
$$\int \frac{dI}{I_0} = 1$$

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songoku
anuttarasammyak said:
I wonder how your ##I_0 =\mathscr{I}_0 \, 2a## plays a role here. Say ak >> 1
$$\int \frac{dI}{I_0} = 1$$
Rearrange ##\int \frac{dI}{I_0} = 1## as $$I_0 = \int dI.$$ Using ##I_0 =\mathscr{I}_0 \, 2a##, this is$$\mathscr{I}_0 \, 2a = \int dI.$$ The LHS equals the total energy per second arriving at a unit length of the slit. The RHS is the total energy per second diffracted at all angles by a unit length of the slit. So, this just represents conservation of energy.

songoku
Your ##\mathscr{I}_0 ## has unit Watt/m^2 and ##I_0## has unit Watt/m. May I interprete that your ##I_0## includes large waste energy shuttered by wall because it has no condition on sideway ?

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anuttarasammyak said:
Your ##\mathscr{I}_0 ## has unit Watt/m^2
Yes, I defined ##\mathscr{I}_0## to be the intensity of the light (Watt/m^2) arriving at any point of the infinitely long slit. So, the amount of energy per second arriving within a unit length of the slit would be ##\mathscr{I}_0## times the width, ##2a##, of the slit: ##\mathscr{I}_0 \, 2a##. This has units Watt/m where the m refers to considering a unit length of the slit.

anuttarasammyak said:
and ##I_0## has unit Watt/m.
Yes. I interpret the Landau and Lifshitz symbol ##I_0## (Watt/m) as also representing the total energy per second arriving at a unit length of the slit. ##\mathscr{I}_0 \, 2a = I_0##.

anuttarasammyak said:
May I interprete that your ##I_0## includes large waste energy shuttered by wall because it has no condition on sideway ?
I'm not sure what this means. Can you rephrase it? ##I_0## includes all of the energy that arrives at the empty space of a unit length of the slit. It does not include any energy arriving at the material wall surrounding the slit opening.

songoku
I agree with you. Let me say my understanding.

$$\frac {a k} {\pi}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d \theta=\frac {2a} {\lambda}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d \theta=\frac {1} {\pi}\frac {\sin^2 (ka\theta)}{(ka\theta)^2} d (ka\theta)$$
expresses distribution according to angle which is normalized to 1. We should multiply input energy ( per second per unit slit length) [watt/m] = 2a [m] * incident light energy (per second per unit area) [watt/m^2] whose dimension is same as ##E^2c##, to it to get the result we need, i.e. energy per slit unit length, per second, per unit angle. So the coefficient is proportional to (2a)^2/##\lambda##.

Re my post #2, Orange line area is double of Blue line area. Thanks.

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songoku and TSny
Thank you very much for all the help and explanation annutarasammyak, vela, Steve4Physics, Tsny

Steve4Physics, anuttarasammyak and TSny

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