Question

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.

Answer 1

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)

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

$\theta = 12.27\°$