How is the width of a slit related to the intensity of light passing through it?

Question

Here's a question I got in my final exam this morning. "If in Young's double-slit experiment setup, the ratio of the intensity of the bright spot to the dark spot is 25:9, what is the ratio of the width of the slits?"

Here's what I did. Since the ratio of intensity at the bright and dark spots is 25:9, the ratio of amplitudes there must 5:3. Which means the amplitude of one wave is 4 times the other.

Now, knowing that the amplitude of light through the wider slit is 4 times the amplitude of light through the narrower slit, how can I determine the ratio of the slits' width?

Answer 1

I hope you know that intensity $(I)$ of light at any point on the screen due to interference in the Young's Double Slit experiment can be given as

$$A^2=I=a_1^2+a_2^2+2a_1a_2\cos{\phi}$$

where $a_1, a_2$ are the amplitudes of the light waves with a constant phase difference of $\phi$, $A$ is the amplitude of the resultant displacement at the point on the screen. For simplicity, we can assume that intensity of the light to be equal to the square of the amplitude as given above.

Thus, $$I_{max}=a_1^2+a_2^2+2a_1a_2(1)=(a_1+a_2)^2$$

$$I_{min}=a_1^2+a_2^2+2a_1a_2(-1)=(a_1-a_2)^2$$

Therefore, $\frac{I_max}{I_min}=\frac{(a_1+a_2)^2}{(a_1-a_2)^2}=\frac{25}{9}$

Thus, $a_1+a_2=5, a_1-a_2=3$

$a_1+(a_1-3)=5=2a_1-3$
Thus, $a_1=8/2=4, a_2=1$

The intensity of light due to a slit (source of light) is directly proportional to the width of the slit. Therefore, if $w_1$ and $w_2$ are widths of the tow slits $S_1$ and $S_2$; $I_1$ and $I_2$ are intensities of light due to the respective slits on the screen, then

$$\frac{w_1}{w_2}=\frac{I_1}{I_2}=\frac{a_1^2}{a_2^2}=\frac{4^2}{1^2}=16$$

Answer 2

The amplitude should be proportional to the width.

In single slit diffraction calculations, the resultant amplitude is obtained by dividing the slit width into a large number of equal segments. For each segment, the amplitude is taken proportionally equal and a constant phase difference is taken as existing between adjacent segments. The resultant amplitude is found by superposition at the point of consideration.

So amplitude should be proportional to slit width. The intensity is proportional to the square of slit width, as intensity is proportional to the square of the amplitude.

Answer 3

actually double width can be imagined as a two slits overlapping with each other. that means the path difference right at the beginning will be zero. Hence amplitudes will add up. So i would say if width is doubled, then amplitude will double. not intensity.