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

The Question:- Page 1Page 1 Page 2Page 2

Note by Rajdeep Dhingra
1 year, 7 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list
  • bulleted
  • list

1. numbered
2. list
  1. numbered
  2. 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"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai−1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 ∑i=13 \sum_{i=1}^3
\sin \theta sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Log in to reply

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

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...