Question

A glass plate 3.45 mm thick, with an index of refraction of 1.60, is placed between a point source of light with wavelength 500 nm (in vacuum) and a screen. The distance from source to screen is 2.05 cm. How many wavelengths are there between the source and the screen?

Answer 1

`4.51* 10^4`

Step-by-step explanation:

We are given that

Thickness of glass plate=3.45 mm=
`3.45* 10^(-3)m`

Using
`1mm=10^(-3) m`

Refractive index=
`n=1.6`

Wavelength in air,
`\lambda_0=500nm=500* 10^(-9) m`

`1 nm=10^(-9) m`

Distance from source to screen=d=2.05 cm=
`2.05* 10^(-2) m`

`1 cm=10^(-2) m`

Wavelength in medium=
`\lambda=(\lambda_0)/(n)`

Using the formula

Wavelength in medium=
`\lambda=(500* 10^(-9))/(1.6)=312.5* 10^(-9) m`

Distance between source and scree except glass plate=
`2.05* 10^(-2)-3.45* 10^(-3)=17.05* 10^(-3) m`

Number of wavelength=
`(distance];in\;air)/(wavelength\;in\;air)+(distance\;in\;glass)/(wavelength\;in\;glass)`

Number of wavelength=
`(17.05* 10^(-3))/(500* 10^(-9))+(3.45* 10^(-3))/(312.5* 10^(-9))`

Number of wavelength=
`4.51* 10^4`