A light ray emits from the origin making an angle 30 ° with the...

On plotting the diagram of given data we get,

Let $Q ( h, 0 )$

$∵$$OP$ reflected by $x + y = 1$

So, the image of $Q$ lies on $y = \dfrac{x}{\sqrt{3}}$

$∴ \dfrac{x - h}{1} = \dfrac{y}{1} = \dfrac{- 2 ( h - 1 )}{2}$

$⇒ x = 1, y = 1 - h$

Since it lies on $y = \dfrac{x}{\sqrt{3}}$

$∴ 1 - h = \dfrac{1}{\sqrt{3}}$

$⇒ h = 1 - \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3} - 1}{\sqrt{3}}$

On rationalising the numerator we get,

$⇒ h = \dfrac{\sqrt{3} - 1}{\sqrt{3}} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$⇒ h = \dfrac{2}{3 + \sqrt{3}}$