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

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

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}`	$\boxed{123}$

Comments

Sort by:

Top Newest

@Josh Silverman @Steven Zheng

Rajdeep Dhingra - 2 years, 10 months ago

Can someone point out the mistake?

Kushal Thaman - 12 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 - 1 year, 7 months 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}`	$\boxed{123}$