Answer
Verified
386.6k+ views
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
\[{d_o} = n{d_g}\]
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
\[{d_o} = n{d_g}\]
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
A block of mass 5 kg is on a rough horizontal surface class 11 physics JEE_Main
A pendulum clock keeps the correct time at 20 degrees class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
The diagram given shows how the net interaction force class 11 physics JEE_Main
Which of the following meet the requirements of Huckel class 11 chemistry JEE_Main