Light bending in gravitational field

Please see the following question \rightarrow link I have solved the first part, but instead of a positive sign, I get a negative one. This is due to that they take potential GMr-\dfrac{GM}{r}. Now in part b of the question, I get the answer by simply differentiating the optical path length with respect to 'closest distance' and I get the required answer, but I don't really understand why. Can you please tell why is the differential of optical path length equal to the angle turned by the light? Or is it just a coincidence? If so, what should be the correct method for solving part(b)?

Note:- Make suitable approximations wherever necessary.

My solution:- My solutionMy solution

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Note by Rajdeep Dhingra
1 year, 7 months ago

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Refractive index has been provided in the problem, hence relativity part is done. @Rajdeep Dhingra Index of refraction provides the speed of light relative to the vaccum speed of light. As I told in quora, it's just Fermat's principle.
You haven't calculuated the path distance correctly, it's not a straight line.

Pawan Goyal - 4 months ago

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