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Byju's Answer

Standard X

Physics

Total Internal 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
$30^{\circ}$

$∠ A O P + ∠ A O N = 90^{\circ}$ (as ON is perpendicular to POQ)
Given, $∠ A O P = 60^{\circ}$
$⟹ ∠ A O N = 90^{\circ} - 60^{\circ}$
$∠ A O N = 30^{\circ}$

By Laws of reflection,
$∠ A O N = ∠ B O N = x$
Thus, $x = 30^{\circ}$

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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). The value of $∠ A O P$ is $60^{\circ}$ . Find the value of the angle of reflection.

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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∘ ∠AOP+∠AON=90∘ (as ON is perpendicular to POQ) Given, ∠AOP=60∘ ⟹∠AON=90∘−60∘​ ∠AON=30∘ By Laws of reflection, ∠AON=∠BON=x Thus, 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.

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

$∠ A O P + ∠ A O N = 90^{\circ}$ (as ON is perpendicular to POQ)
Given, $∠ A O P = 60^{\circ}$
$⟹ ∠ A O N = 90^{\circ} - 60^{\circ}$
$∠ A O N = 30^{\circ}$

By Laws of reflection,
$∠ A O N = ∠ B O N = x$
Thus, $x = 30^{\circ}$