Question

The optical path of the monochromatic light is the same if it goes through 4.0cm of glass or 4.5cm of water. If the refractive index of glass is 1.53, the refractive index is
A. 1.30
B. 1.36
C. 1.42
D. 1.46

Answer 1

Hint Optical path length has a magnitude given by the product of geometric length, followed by light in any medium and its refractive index. Mathematically optical length is given by $OPL = n \times s$ where n is the refractive index of the medium and s is the length of the medium through which light travelled.

Complete step by step answer:
The path followed by light in an optical medium is often called an optical path. The optical path of the light while travelling between two different media is affected by various factors such as reflection, (total internal reflection), refraction, absorption , dispersion of light etc.
Optical path is a very useful concept when we consider phase difference between two coherent waves that are superposed after travelling different distances ${X_1}$ and ${X_2}$ in a medium of refractive index n. Then
$\dfrac{\theta }{{2 \times \pi }} = \dfrac{{{X_2} - {X_1}}}{{{\lambda _n}}}$
 Here ${\lambda _n}$ is the wavelength of the light that is medium and $\theta $ is the phase difference.
Using optical path differences we can calculate the phase difference between two coherent waves travelling in mediums.
When a wave travels from one medium to another there is no change in wave frequency. But the other factors like wave speed and wavelength vary from one medium to another.
Here in the glass water interface it is said that their optical path are same
${\mu _1}{x_1} = {u_2}{x_2}$
i .e they have zero optical path difference Substituting the value of ${X_1} = 4cm$ , ${X_2} = 4.5cm$ , ${\mu _1} = 1 \cdot 5$
We get $4 \times 1.53 = 4.5 \times {X_2}$
                   ${X_2} = \dfrac{{6.12}}{{4.5}}$
                   ${X_2} = 1.36$

Note: When optical path differences from one whole wavelength their phase difference will be $2 \times \pi $. In the earlier given equation $\dfrac{{{X_2} - {X_1}}}{{{\lambda _n}}}$ can be replaced as $\dfrac{{n({X_2} - {X_1})}}{\lambda }$ where the numerator show optical path difference and denominator shows the wavelength of light in vacuum.