Question

Blue light of wavelength 475 nm passes through an interference grating with a slit spacing of 0.002 mm and makes an interference pattern on the wall. How many bright fringes will be seen?

Answer 1

To solve this problem we apply the principle of constructive superposition (Diffraction from a single slit), for which we have

$dsin\theta = m\lambda$

Here,

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

d = Slit separation

$\lambda$ = Wavelength

Rearranging for n,

$n = \frac{dsin\theta}{\lambda}$

Maximum diffraction possible is when the value of sin\theta is maximum.

Replacing we have tat,

$n = \frac{(0.002*10^{-3})sin(90)}{475*10^{-9}m}$

$n = 4.211$

When diffraction takes place, the fringes are formed on both sides of the center of pattern.

These fringes are dark and brigth fringes formed alternatively in a triangular pattern.

The total number of brigth fringes will be equal to double of the fringes formed on one side of the center of the pattern.

The number of the fringes then is

$n' = 2n$

$n ' = 2(4.211)$

$n' = 8.422 \approx 9$

Therefore the total number of brigth is 9