Optical Geometry : "Fun with Laser Light"

Consider an rectangular glass board $ABCD$ , with surfaces $AB$ , $BC$ , $CD$ are perfectly silvered. An small laser light is set in the corner $A$ (negligible dimensions) . Now just for fun , this laser light is set in such a way so that light ray first travel from A making an angle $\theta$ with side $AD$ and then totally reflected by surfaces $DC$ , then $CB$ , then $BA$ and finally refracted out from surface $AD$ . If during this event , an quadrilateral and an triangle are traced out as shown & their point of intersection is $P$ , Such that Area of triangle( $APT$ ) is equal to that of quadrilateral( $PQRS$ ).

Given that :

$AB=CD=2(BC)=2(AD)$

$\text{Laws of reflection holds perfectly}$ .

Then find that particular angle $\theta$ in degrees or just simply prove that $\tan { \theta } =\cfrac { a-\sqrt { b } }{ c }$ and find $a+b+c$ , for which such event is possible.

#Geometry #FUN #Optical-geometry #Laser-light-fun

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

I get Tan(theta)= 4-√8 Can someone confirm

Suhas Sheikh - 2 years, 11 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}`	$\boxed{123}$