# A glass plate 2.50 mm thick, with an index of

A glass plate 2.50 mm thick, with an index of refraction of 1.40, is placed between a point source of light with wavelength 540 nm (in vacuum) and a screen. The distance from source to screen is 1.80 cm. How many wavelengths are there between the source and the screen?
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Jimmy Macias
Step 1
The number of wavelength in a distance d can be calculated using the following formula:
$\text{Number of wavelengths}=\frac{dis\mathrm{tan}ce}{wave\le n>h}$
$=\frac{d}{\lambda }$
And wavelength $\left(\lambda \right)$ in a medium having a index of refraction n will be:
$\frac{{\lambda }_{0}}{n}$
(${\lambda }_{0}$ is wavelength in air, n is index of refraction)
Step 2
Wavelength in the glass plate will be:
$\lambda =\frac{540}{1.4}$
$⇒\lambda =385.7nm$
Length between source to screen is 1.8 cm, glass plate is of 2.5 mm thickness. Distance between source and screen excluding glass plate is 1.55 cm $\left(1.88-0.25\right)$.. So the number of wavelength will be.
$Number=\frac{\text{distance in air}}{\text{wavelength in air}}+\frac{\text{distance in glass}}{\text{wavelength in glass}}$
$=\frac{1.55×{10}^{-2}}{540×{10}^{-9}}+\frac{2.5×{10}^{-3}}{385.7×{10}^{-9}}$
$=2.87×{10}^{4}+0.64×{10}^{4}$
$=3.51×{10}^{4}$
Hence, the total number of wavelengths will be $3.51×{10}^{4}$.

trisanualb6
$N=0.194$
Explanation:
Given:
- refractive index of the glass plate, $n=1.4$
- thickness of the glass plate, $x=2.5mm$
- wavelength of the source light, $\lambda =18mm$
We know that the refractive index of a medium is given as:
$n=\frac{\text{wavelength of light in the air or vacuum}}{\text{wavelength of light in the medium}}$
$n=\frac{\lambda }{{\lambda }^{\prime }}$
$1.4=\frac{18}{{\lambda }^{\prime }}$
Hence the no. of wavelengths in the glass:
$N=\frac{x}{{\lambda }^{\prime }}$
$N=\frac{2.5}{12.857}$
$N=0.194$