Question

Explain what is optical path length. How is it different from actual path length?

Answer 1

Hint: Optical path length can be given as the product of the refractive index and the distance which would have travelled by light at a particular time. The actual path length is the distance
travelled by light in a medium.

Formula used:
We will be using the following formulae; \[n = \dfrac{{{v_v}}}{{{v_m}}}\] where \[n\] is the index of refraction of a particular medium, \[{v_v}\] is the speed of light vacuum, \[{v_m}\] is the speed of light in that medium. \[{d_o} = n \times {d_g}\] where \[{d_o}\] is the optical path length, \[n\] is the refractive index of the medium and \[{d_g}\] is the actual path length or the geometric length.

Complete step by step solution:
Generally optical path length and actual path length are related but are not identical.
The optical path length of light in a medium can actually be defined as the length or distance in which the light would have travelled in the same time if it were travelling in a vacuum. This statement means that if a light travels through a particular medium, it would have travelled a particular length or distance within a particular time, but the optical path length is the length or distance that light waves would have travelled if it were not travelling in a medium but in vacuum.
Generally, the optical length can be given as
\[{d_o} = n \times {d_g}\] where \[{d_o}\] is the optical path length, \[n\] is the refractive index of the medium and \[{d_g}\] is the actual path length or the geometric.
Actual path length or Geometric length is simply the real length travelled by the light in the medium.

Note: For clarity, we can prove the formula for the optical path length as follows:
The distance travelled by light in the medium is
\[{d_g} = {v_m} \times t\]
but now, the distance the length would have travelled in vacuum in the same time would be
\[{d_o} = {v_v} \times t\]
\[ \Rightarrow t = \dfrac{{{d_o}}}{{{v_v}}}\]
Inserting into \[{d_g} = {v_m} \times t\]
\[{d_g} = {v_m} \times \dfrac{{{d_o}}}{{{v_v}}}\]
By making \[{d_o}\] subject of the formula,
\[{d_o} = \dfrac{{{v_v}}}{{{v_m}}}{d_g}\]
Now recalling that, \[n = \dfrac{{{v_v}}}{{{v_m}}}\] where \[n\] is the index of refraction of a particular medium, \[{v_v}\] is the speed of light vacuum, \[{v_m}\] is the speed of light in that medium, then
\[{d_o} = n{d_g}\]