Question

Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.15°. What is the width of the slit?

Answer 1

`d =4.77* 10^(-5)\ m`

Step-by-step explanation:

given

wavelength of light λ = 479 nm

= 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°

using

condition for diffraction minimum ,

d sinθ = m λ

for first minimum m = 1

d sinθ = λ

therefore ,

slit width

`d =(\lambda)/(sin\theta)`

`d =(479* 10^(-9))/(sin 0.575^0)`

`d =(479* 10^(-9))/(0.01)`

`d =4.77* 10^(-5)\ m`

hence, the width of the slit is equal to
`d =4.77* 10^(-5)\ m`