Question

A glass plate 2.25 mm thick, with an index of refraction of 1.80, is placed between a point source of light with wavelength 620 nm (in vacuum) and a screen. The distance from source to screen is 1.50 cm. How many wavelengths are there between the source and the screen?

Answer 1

Answer:

There are 2.71x10⁴ wavelengths between the source and the screen.

Explanation:

The number of wavelengths (N) can be calculated as follows:

$N = \frac{d_{g}}{\lambda_{g}} + \frac{d_{a}}{\lambda_{a}}$    

Where:

$d_{g}$: is the distance in glass = 2.25 mm

$d_{a}$: is the distance in air = 1.50 cm - 0.225 cm = 1.275 cm

$\lambda_{g}$: is the wavelength in glass

$\lambda_{a}$: is the wavelength in air = 620 nm

To find the wavelength in glass we need to use the following equation:

$n_{g}*\lambda_{g} = n_{a}*\lambda_{a}$

Where:

$n_{g}$: is the refraction index of glass = 1.80

$n_{a}$: is the refraction index of air = 1

$\lambda_{g} = \frac{\lambda_{a}}{n_{g}} = \frac{620 nm}{1.80} = 344.4 nm$

Hence, the number of wavelengths is:

$N = \frac{d_{g}}{\lambda_{g}} + \frac{d_{a}}{\lambda_{a}}$

$N = \frac{2.25 \cdot 10^{-3} m}{344.4 \cdot 10^{-9} m} + \frac{1.275 \cdot 10^{-2} m}{620 \cdot 10^{-9} m}$                                    

$N = 2.71 \cdot 10^{4}$

Therefore, there are 2.71x10⁴ wavelengths between the source and the screen.

I hope it helps you!