Question

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength 557 nm falls on a double slit whose slit separation is 3.36 × 10^-6 m?

Answer 1

It can be solved with a concept of path difference.

Path difference

nY = d * sin(theta) - equation 1

Where n is a positive real number and Y is the wavelength of the light.

d is the distance between slits.

For upto last bright fringe on the screen the angle wil be 90° .

So equation 1

n*Y = d* (1) as sin90° = 1

n = d/Y

Now n will give the count of total bright fringes on one side.

n = 3.36 x 10 -6/ 557 x 10 -9

n = 6.03

n= can't be a decimal number.

so the n= 6

6 on each side

total 6*2 = 12 on both sides

and if we also Include central bright fringe then 12 + 1 = 13.