fbpx

GATE 2021 Path Difference Geology Question! Is it correct?

The following question appeared in the 2021 Geology and Geophysics GATE exam: Light passes through two media with refractive indices of 1.75 and 1.55, respectively. The thickness of both the media is 30 µm. The resultant path difference of the yellow light component (λ = 589 nm) is _________ µm. (Take λ = 3.141) [round off to one decimal places]. This question left students confusing as the answer provided in the official (and final) answer key is 64 µm. Let us see what answer we get when we solve this question. Path difference: Since the two media have different refractive ...

The following question appeared in the 2021 Geology and Geophysics GATE exam:

Light passes through two media with refractive indices of 1.75 and 1.55, respectively. The thickness of both the media is 30 µm. The resultant path difference of the yellow light component (λ = 589 nm) is _________ µm. (Take λ = 3.141) [round off to one decimal places].

This question left students confusing as the answer provided in the official (and final) answer key is 64 µm.

Let us see what answer we get when we solve this question.

Path difference: Since the two media have different refractive indices, light travels at different velocities in these two media. Light travels faster in the second medium (RI=1.55) and slower in the first medium (RI=1.75). The faster ray will emerge out of the medium before the slower ray. By the time the slower ray emerges out, the faster ray would have travelled some distance in the surrounding medium. This distance is equal to the path difference. The path difference can be calculated using the formula:

∆x = (RI1 – RI2)t

Where,

  • ∆x = path difference
  • RI1 = refractive index of the first medium = 1.75
  • RI2 = refractive index of the second medium = 1.55
  • ht = thickness = 30 µm

Now you must be wondering, why does the question mention the wavelength and π, which are not required in the calculation of path difference. We will answer that in a while, but first, let us calculate the path difference using the formula above.

∆x = (1.75 – 1.55)30 µm = 0.2 x 30 µm = 6 µm

∆x = 6 nm

Check out the formula for path difference (also known as retardation) from Tulane notes.

Clearly this does not match the answer provided in the official answer key.

Now, there is another quantity closely related to path difference. It is called the Phase difference.

A path difference of one wavelength (λ) is equal to a phase difference of 2π. In general path difference can be related to phase difference using the formula:

Phase difference = ∆ϕ = 2π∆x/λ

You can see here that to calculate the phase difference you need π andλ both. Using the values provided in the question we get:

∆ϕ = 2×3.141×6000/589 = 63.99 radians

Rounding off to 1 decimal place we get a phase difference of 64.0 radians. The numerical value (but not the units) here matches the answer provided in the official key.

It appears that GATE has committed an error in this question. The asked to calculate the path difference, but have provided the answer for phase difference.

Chat with us!