PHYSICS 3.... Light of wavelength of 480nm is incident on a single slit opening

Question

PHYSICS 3....

Light of wavelength of 480nm is incident on a single slit opening with a width of 6micrometers.  If the screen is 1m from the opening,  A) how wide is the central maximum?  B) If instead we were using sound with a frequency of 100 Hz, what is the maximum width of the door in order to hear an interference pattern?  (You can assume that the speed of sound is 343 m/s)

I know that (asin@=m*wavelength) but I'm confused on what m should i use for the first minima or maxima . Please Explain ..

Thank You

Explanation / Answer

Angular width of central maximum is

= 2 lamda/ W

= 2 ( 480 nm)/ 6 * 10 ^-6 m

= 0.160 m

(B)

wavelength of sound wave is

wavelength = v/f = 343 m/s/ 100 Hz = 3.43 m

condition for interference maximum is

d sin theta = m Lamda

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