Question

23. Two slits are separated by 0.180 mm. An interference pattern is formed on a screen 80.0 cm away by 656.3-nm light.

Calculate the fraction of the maximum intensity a distance y 5 0.600 cm away from the central maximum.

Answer 1

The fraction of the maximum intensity a distance y 5 0.600 cm away from the central maximum is 0.97.

Step-by-step explanation:

d = 0.180 mm = 1.8 x 10⁻⁴ m

L = 80 cm = 0.8 m

λ = 656.3 nm = 6.563 x 10⁻⁷ m

y = 0.6 cm = 6 x 10⁻³ m

we know that the equation of the intensity for two slits interference equation is

I =
`I_(max)` cos²(πdy/λL)

where

I = intensity

`I_(max)` = maximum intensity

d = distance of two slits

y = the distance of the interference with the central

L = the distance of screen and the slits

λ = wavelength

so,

I =
`I_(max)` cos²(πdy/λL)

I/
`I_(max)` = cos²(πdy/λL)

= cos²[π (1.8 x 10⁻⁴) (6 x 10⁻³)/(6.563 x 10⁻⁷) (0.8 )]

= 0.97