Question

In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? 214 nm 204 nm 224 nm 100 nm 234 nm

Answer 1

The value is
`\lambda  = 214.3  \ nm`

Step-by-step explanation:

From the question we are told that

The slit separation is
`d =  3.00 * 10^(-5) m`

The distance of the screen is
`D =   2.00\ m`

The order of fringe is n = 7

The path difference is
`y =  10.0 \ cm  =  0.1 \  m`

Generally the path difference is mathematically represented as

`y =  (n *  \lambda  *  D)/( d)`

=>
`0.1 =  (7 *  \lambda  * 2.00 )/( 3.00 * 10^(-5))`

=>
`\lambda  =  (0.1 *3.00 * 10^(-5) )/(7 * 2.00 )`

=>
`\lambda  =  (0.1 *3.00 * 10^(-5) )/(7 * 2.00 )`

=>
`\lambda  = 2.143 *10^(-7) \  m`

=>
`\lambda  = 214.3  \ nm`