A ray of light travelling along the line x $$+$$ $$3y$$ $$=$$ $$5$$ is incident on the $$x-axis$$ and after refraction it enters the other side of the $$x-axis$$ by turning π$$6away$$ from the $$x-axis$$. The equation of the line along which the refracted ray travels is

Question

A ray of light travelling along the line x $$+$$ $$3y$$ $$=$$ $$5$$ is incident on the $$x-axis$$ and after refraction it enters the other side of the $$x-axis$$ by turning π$$6away$$ from the $$x-axis$$. The equation of the line along which the refracted ray travels is

Infinity Learn · Accepted Answer

The refracted ray passes through the point (5$$, $$0) and makes an angle $$120$$º with positive direction of $$x-axis$$∴ The equation of the refracted ray is (y$$ – $$0) $$=$$ $$tan$$ $$120$$º (x$$ – $$5)⇒$$y=-3(x-5)$$  or  $$3x+y-53=0$$

A ray of light travelling along the line x + 3y = 5 is incident on the x-axis and after refraction it enters the other side of the x-axis by turning $\frac{π}{6}$ away from the x-axis. The equation of the line along which the refracted ray travels is

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A ray of light travelling along the line x + 3y = 5 is incident on the x-axis and after refraction it enters the other side of the x-axis by turning π6away from the x-axis. The equation of the line along which the refracted ray travels is

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Ready to Test Your Skills?

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A ray of light travelling along the line x + 3y = 5 is incident on the x-axis and after refraction it enters the other side of the x-axis by turning $\frac{π}{6}$ away from the x-axis. The equation of the line along which the refracted ray travels is