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 $-\dfrac{GM}{r}$ . 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 solution

The Question:- Page 1 Page 2

Easy Math Editor

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Comments

@Josh Silverman @Steven Zheng

Rajdeep Dhingra - 3 years, 5 months ago

Can someone point out the mistake?

Kushal Thaman - 1 year, 6 months ago

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 - 2 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$