Optics Diffraction Problems: Finding Maximum Intensity Angles and Slit Widths

[franz32]

Specifically, diffraction.. here are the insights of the problems...

1. For a radio station operating at a frequency of 103.3 MHz has two identical dipole antennae 1000m apart, transmitting in a phase. At distances much greater than 1000m, at what angles is the intensity at a maximum?

I believe I should use d sin 'theta' = m 'lambda' and find 'theta'.

Here's my problem... I will only get the 'theta' by substitution of necessary value to the formula, but how will I find the angleS? What should I do with the formula? Make multiples of 1000m and m =2,3,...?

2. Consider a straight black line drawn on a piece of white paper. After taking a film photo-graph of this arrangement, the resulting negative shows a thin, transparent line surrounded by opaque black. A laser, of 633 nm wavelength, strikes this negative, and the pattern is observed on a screen 60 m away. The distance between the central bright fringe and the first minimum(dark spot) is 32 mm. Determine the width of the slit on the negative.

Insights: Can I use y = Rn 'lambda' / a where R is distance from slit to screen, y is distance between minima? So, the unknown is the 'a'.

with 'lambda' = 633 nm; R = 60 m;

Question: SHould my n be equal to 1? How about my y? 32mm.

3. Coherent light from a mercury arc-discharge lamp is passed through a filter that blocks everything except for one spectral line in the green region of the spectrum. This light then falls on two slits separated by 600 mm. The resulting interference pattern is projected on a screen 2.5 mm away, and the adjacent bright lines are each separated by 5 mm. Determine the wavelength of the light used.

y = Rn 'lambda' / a​

My a is = 600mm, R = 2.5mm, 'lambda' is unknown... and y = 5mm? what should my n be?

 

[M98Ranger]

1. To find the maximum intensity angles, you can use the formula d sin 'theta' = m 'lambda', where d is the distance between the antennae (1000m in this case), 'theta' is the angle at which the maximum intensity occurs, m is the order of the maximum (1, 2, 3...), and 'lambda' is the wavelength of the radio waves (calculated using the frequency of 103.3 MHz). By substituting different values for m, you can find the corresponding angles at which the intensity is at a maximum.

2. For the second problem, you can use the formula y = Rn 'lambda' / a, where y is the distance between the central bright fringe and the first minimum, R is the distance between the negative and the screen (60m), 'lambda' is the wavelength of the laser (633nm), and a is the width of the slit on the negative (unknown). By solving for a, you can determine the width of the slit. The value of n will depend on which minimum you are considering (first, second, third, etc.). In this case, since the distance between the central bright fringe and the first minimum is given, you can use n=1. The value of y is 32 mm, but it would be better to convert it to meters (0.032m) for consistency with the other units in the equation.

3. In the third problem, you can use the same formula y = Rn 'lambda' / a, where y is the distance between adjacent bright lines (5mm), R is the distance between the slits and the screen (2.5mm), and 'lambda' is the wavelength of the light (unknown). The value of n will depend on which bright line you are considering (first, second, third, etc.). In this case, since the adjacent bright lines are separated by 5mm, you can use n=1. The value of a is given as 600mm, but it would be better to convert it to meters (0.6m) for consistency with the other units in the equation. By solving for 'lambda', you can determine the wavelength of the light used.