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How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?

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7 votes

Answer:

6

Step-by-step explanation:

We are given that


\theta=2.12^(\circ)

Slid width,a=0.110 mm=
0.11* 10^(-3) m


1mm=10^(-3) m

Wavelength,
\lambda=582 nm=582* 10^(-9) m


1nm=10^(-9) m

We have to find the number of diffraction maxima are contained in a region of the Fraunhofer single-slit pattern.


asin\theta=(2N+1)/(2)\lambda

Using the formula


0.11* 10^(-3)sin(2.12)=(2N+1)/(2)(582* 10^(-9))


2N+1=(0.11* 10^(-3)sin(2.12)* 2)/(582* 10^(-9))


2N+1=13.98


2N=13.98-1=12.98


N=(12.98)/(2)\approx 6

Hence, 6 diffraction maxima are contained in a region of the Fraunhofer single-slit pattern

User PEWColina
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