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Microwave of an unknown wavelength is incident on single slit of width cm_ The angular width of the central peak is found to be 290 Find the wavelength (in cm)_cm...

Question

Microwave of an unknown wavelength is incident on single slit of width cm_ The angular width of the central peak is found to be 290 Find the wavelength (in cm)_cm

microwave of an unknown wavelength is incident on single slit of width cm_ The angular width of the central peak is found to be 290 Find the wavelength (in cm)_ cm



Answers

A microwave of an unknown wavelength is incident on a single slit of width $6 \mathrm{cm} .$ The angular width of the central peak is found to be $25^{\circ} .$ Find the wavelength.

If the minimum occurs at 10 a.m. then scientist, I am multiplied by the seat with a equals m lambda. Now you're Didn't we ducked? We're dealing with m equals one. So the equation becomes there's no one equals Sign in the worst land over a equals signing Worse 6 33 Change the minus nine meter and the sleep opening is given by 25 micron. Two point for my crone. So 2.4 times 10 to the native six meter. Now, if this is the horizontal line here we have Maxima. Andi, then we have minimal Over here. This angle is the better one we discussed. So angle between minimal on this side, Minimal on the topside. To the minimum bottom side is Tedder. So the required angle is to terror. We just wise that it was twice that if we put the numbers in, then that comes out to be 2 89 degree

Mhm. For a single set diffraction. The position of dark fringe, Yeah, position of a dark fringe measured from the central maximum is why. And that's given by this question why calls eight times lambda times L over split with. So he is an integer. That's the index of the dark fringes starts from one lambda is the wavelength, is the distance on the sled to the screen. And then we have to sleep with. We plug in our numbers for the information. Given the question 0.67 m. For why P equals one lambda is the 2.8 centimeters convert to meters at least one m. And we're trying to find the slip with. So we rearrange the question sell for the slip with they get 0.42 m. That's 4.2 centimeters.

So here we are dealing with single slit diffraction. So in think of the diffraction to find the angler position. That is the angle between the dark fringe and the center. Um And the measured from the a single fit itself is given by this equation. Theta is the angle equal speed time, slander over to sleep with. So P is just an integer that starts from one and goes up lumber. Is the wavelength liquid? Is the object is to give and sleep with. So here we are looking at the central maximum. Okay so we plug in equals one, then plug in lambda in meters, 500 nanometers in meters. And the flip with um 1.5 micrometers also convert to meters. And the theater that began building radiance. So the calculation gives us 0.333 radiance. And you can't wait to decrease. You get 19.1°. So this is the angle at which the first dark fringes located measured from the center Um on one side of the central maximum. But the question is asking about the angular width of the central maximum, which means we have measured from one side to the other side. So That basically means you have to double the angle from the that they just found, and that gives us 38.2°.

So this is problem 24.21. This problem have a monochromatic light with weapons 680 millimeters and just like is going through a single slip. Okay. And then we will observe a diffraction better. And, ah, the weight of this letters opened or for 25 millimeters. And ah, the question is asking us to find the wits off the central Ah, peak here. Okay, so we know that, um, the science data equal to Lambda for a single state experiment like this and ah, fatal here is actually this angle and we'll hear from the blue line here from them minimum off the peak to the maximum of the big this angle data. So, actually, the wits off the full peak will be to theater. This angle, which I call for here, will be so safe, Right? So in order to find faith, huh? We can just you this relationship be scientist able to London because we have lambda and ah Dee So we can find data. It will be just sign of Earth off Lambda Over D, right. And so alpha will be two time sign of theirs, Lambda Over the which can be written as to sign Evers of, Then I'll substitute with my values here. So 6 80 time Stemple minus nine meters over D, which is 0.4 25 beaters, which will give you an angle, which is one with a the four degrees, unless will be the width off the central maximum. So I have an important thing to observe here that data just half off the angular with off the central maximum. It's not the full with off the central maximum.


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