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ExtremeBDS [4]
4 years ago
15

Monochromatic light of wavelength 491 nm illuminates two parallel narrow slits 6.93 μm apart. Calculate the angular deviation of

the third-order (for m = 3) bright fringe (a) in radians and (b) in degrees.
Physics
1 answer:
Papessa [141]4 years ago
3 0

To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,

dsin\theta = m\lambda

Here,

d = Distance between slits

m = Any integer which represent the order number or the number of repetition of the spectrum

\lambda = Wavelength

\theta = Angular deviation

Replacing with our values we have,

(6.93*10^{-6}) sin\theta = (3)(491*10^{-9})

\theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })

Part A)

\theta = 0.2141rad

PART B)

\theta = 0.2141rad(\frac{360\°}{2\pi rad})

\theta =  12.27\°

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