Request a call back

# JEE Class main Answered

Question is in the image attached
Asked by swayamagarwal2114 | 15 Jul, 2022, 09:18: AM
Let P is point of luminous source. Light ray from P incident on A is getting reflected first and
then at B is getting reflected second time and then reaches the source point P as shown in figure .

a is the distance from P to centre O and b is the distance from centre to the point
where reflected ray meets the diamenter through P .

Let us use the property of triangle as applied to ΔPQR as shown in figure.

If the line PS divides the base QR in the ration m : n and makes angle α , β and γ as shown in figure ,

then we have ,  ( m + n ) cotγ = m cotα - n cotβ

By using this property to ΔPAB as shown in left side of figure, we get

( a + b ) cotφ = (a-b) cotθ

it can be shown that φ = 90 - θ , hence cotφ = cot(90-θ) = tanθ

tan2θ = (a-b) / a+b

Answered by Thiyagarajan K | 15 Jul, 2022, 07:42: PM
JEE main - Physics
Asked by pranav10022007 | 12 May, 2024, 10:05: AM
JEE main - Physics
Asked by sreelekhachakali2006 | 15 Mar, 2024, 10:35: AM
JEE main - Physics
Asked by ramanjaneyuluoguru | 22 Jan, 2024, 12:30: PM
JEE main - Physics
Asked by kumarpurshotam451 | 29 Sep, 2023, 03:46: PM
JEE main - Physics
Asked by kumarpurshotam451 | 27 Sep, 2023, 04:57: PM
JEE main - Physics
Asked by swayamagarwal2114 | 15 Jul, 2022, 09:19: AM