Question

The wavelength of light in a vacuum is five thousand angstroms. When it travels normally through the diamond of thickness one millimeter, the number of waves of light in one millimeter of the diamond is:(Refractive index of diamond $ = 2.417$)
A. $4732$ waves
B. $4932$ waves
C. $4834$ waves
D. $3826$ waves

Answer 1

Hint: In this question first do the conversion of units i.e. wavelength which is given in angstrom convert it into meter by multiplying it by ${10^{ - 10}}$ and also thickness which is given in millimeter is taken as $1 \times {10^{ - 3}}$ meter.

Complete step-by-step solution:
Wavelength is a property of length which is the distance between the two successive waves or crests of light waves. The refractive index is the ratio of the velocity of light in a vacuum to the velocity of light in the medium. The optical path is a product of the refracting index of the medium and the geometrical length of the path. Hence optical path=$\mu \times d$. And also it can be written as a product of the number of waves $ \times \lambda $. Hence we find that the optical path is equal to the number of waves $ \times \lambda = \mu \times d$.
 So here is the formula, in which we find the relationship between the number of waves, refracting index, and the wavelength.

Formula used:
Number of waves= $\dfrac{{\mu \times d}}{\lambda }$
where $\mu $ is a refractive index, $d$ is the thickness of diamond and $\lambda $ is the wavelength of light.
Given that,
Refractive index of diamond i.e. $\mu = 2.417$
Wavelength of light in vacuum i.e.$\lambda = 5000{{\rm A}^0} = 5000 \times {10^{ - 10}}m$
Thickness of diamond $ \Rightarrow 1mm = 1 \times {10^{ - 3}}m$
Therefore the number of waves of light in one millimetre of diamond is:
  $
   \Rightarrow \dfrac{{\mu \times d}}{\lambda } = \dfrac{{2.417 \times {{10}^{ - 3}}}}{{5000 \times {{10}^{ - 10}}}} = \dfrac{{2417 \times {{10}^{ - 6}}}}{{0.5 \times {{10}^{ - 6}}}} \\
   = \dfrac{{2417}}{{0.5}} = 4834 \\
$
Hence the correct option is C. The number of waves of light is $4834$ waves.

Note: In this question after the conversion of units we applied the formula where the refractive index of the diamond, the wavelength of light in vacuum, and thickness of the diamond all are given therefore after putting the values we got the result as $4834$ number of waves.