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Standard VIII

Physics

Laws of Reflection

A ray OA is i...

Question

A ray OA is incident upon a smooth surface POQ. It is reflected back along OB, a normal ON is drawn perpendicular to the surface at O (the point of incidence) and the value of $∠ A O P$ is $60^{\circ}$ . Find the value of x.

$30^{\circ}$

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$40^{\circ}$

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$50^{\circ}$

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$60^{\circ}$

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Solution

The correct option is A
$30^{\circ}$

Given, $∠ A O P = 60^{\circ}$

As ON is perpendicular to POQ
$∠ A O P + ∠ A O N = 90^{\circ}$

$⟹ ∠ A O N = 90^{\circ} - 60^{\circ}$
$∠ A O N = 30^{\circ}$

By laws of reflection,
Angle of incidence = Angle of reflection
$⟹ ∠ A O N = ∠ B O N = x$
Thus,
$A n g l e o f r e f l e c t i o n x = 30^{\circ}$

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A ray OA is incident upon a smooth surface POQ. It is reflected back along OB, a normal ON is drawn perpendicular to the surface at O (the point of incidence) and the value of ∠AOP is 60∘. Find the value of x.

The correct option is A 30∘ Given, ∠AOP=60∘ As ON is perpendicular to POQ ∠AOP+∠AON=90∘ ⟹∠AON=90∘−60∘​ ∠AON=30∘ By laws of reflection, Angle of incidence = Angle of reflection ⟹ ∠AON=∠BON=x Thus, Angle of reflection x=30∘

A ray OA is incident upon a smooth surface POQ. It is reflected back along OB, a normal ON is drawn perpendicular to the surface at O (the point of incidence) and the value of $∠ A O P$ is $60^{\circ}$ . Find the value of x.