Question

A Young’s double-slit apparatus is set up where a screen is positioned 0.80 m from the double slits. If the distance between alternating bright fringes is 0.95 cm, and the light source has a wavelength of 580 nm, what is the separation of the double slits?

Answer 1

Answer: 0.0488 mm

Step-by-step explanation:

Given

The position of the screen is
`D=0.80\ m`

Distance between alternating bright fringes is
`0.95\ cm`

The wavelength of the light source is
`\lambda =580\ nm`

Distance between successive bright fringes is
`(\lambda D)/(d)`

Distance between alternating bright fringes is half of the distance between successive fringes i.e.
`(\lambda D)/(2d)`

Insert the values for alternating fringes

`\Rightarrow 0.95* 10^(-2)=(580* 10^(-9)* 0.8)/(d)\quad [\text{d=separation of the doubleslits}]\\\\\Rightarrow d=(464* 10^(-9))/(0.95* 10^(-2))\\\\\Rightarrow d=488.42* 10^(-7)\ m\\\Rightarrow d=0.0488\ mm`