Question

Two slits are illuminated by a 363 nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is 14.9 ◦ . If the screen is 170 cm from the double-slit, how far apart is this bright band from the central peak

Answer 1

y = 0.44 m

Step-by-step explanation:

As we know that path difference on the screen is given as

`\Delta x = (yd)/(L)`

now for constructive interference we know that

path difference = integral multiple of wavelength

so we have

`N\lambda = (yd)/(L)`

now for 4th maximum on the screen we can say

`y = (N\lambda L)/(d)`

here N = 4

L = 170 cm = 1.70 m

`\lambda = 363 nm`

also we know

path difference on screen = d
`sin\theta`

`4\lambda = dsin14.9`

`d = (4\lambda)/(sin14.9)`

now we have

`y = (4(\lambda)(1.70))/((4\lambda)/(sin14.9))`

`y = 0.44 m`