Question

Two slits are separated by 0.380 mm. A beam of 550-nm light strikes the slits, producing an interference pattern. Determine the number of maxima observed in the angular range

Answer 1

Answer:

total maximum on screen = 1381

Explanation:

In the interference pattern we know that path difference for the position of maximum on the screen is given by

$\Delta L = d sin\theta$

here we know that

$\theta$ = angular position of maximum on screen

d = distance between two slits

$\Delta L = (0.380 mm)sin\theta$

here we know that for all maximum positions

$\Delta L = N\lambda$

now plug in all values

$(0.380 \times 10^{-3})sin\theta = N(550 \times 10^{-9})$

here we have

$sin\theta = 1.45 \times 10^{-3}N$

now we know that

$sin\theta < 1$

$1.45 \times 10^{-3} N < 1$

$N < 690.9$

so total number of maximum on screen is

$N = 690 + 690 + 1 = 1381$