A ray of light is incident on a convex mirror | vedclass

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(B) Let the incident vector be $\vec{v}_i = 3\hat i + 4\hat j + 12\hat k$. The normal vector is $\vec{n} = 3\hat i + 4\hat j$. First,find the unit nor

Vedclass · Accepted Answer

$\frac{1}{13}(-3\hat i - 4\hat j + 12\hat k)$

Vedclass · Answer

(B) Let the incident vector be $\vec{v}_i = 3\hat i + 4\hat j + 12\hat k$. The normal vector is $\vec{n} = 3\hat i + 4\hat j$. First,find the unit normal vector $\hat{n} = \frac{3\hat i + 4\hat j}{\sqrt{3^2 + 4^2}} = \frac{3\hat i + 4\hat j}{5}$. The incident ray can be decomposed into components parallel and perpendicular to the normal. The component of $\vec{v}_i$ parallel to $\hat{n}$ is $\vec{v}_{i\parallel} = (\vec{v}_i \cdot \hat{n})\hat{n} = \left( \frac{3(3) + 4(4) + 12(0)}{5} \right) \hat{n} = \frac{25}{5} \hat{n} = 5 \hat{n} = 3\hat i + 4\hat j$. The component of $\vec{v}_i$ perpendicular to $\hat{n}$ is $\vec{v}_{i\perp} = \vec{v}_i - \vec{v}_{i\parallel} = (3\hat i + 4\hat j + 12\hat k) - (3\hat i + 4\hat j) = 12\hat k$. Upon reflection,the parallel component reverses direction,while the perpendicular component remains unchanged. Thus,the reflected vector $\vec{v}_r = -\vec{v}_{i\parallel} + \vec{v}_{i\perp} = -(3\hat i + 4\hat j) + 12\hat k = -3\hat i - 4\hat j + 12\hat k$. The unit vector along the reflected ray is $\hat{v}_r = \frac{-3\hat i - 4\hat j + 12\hat k}{|\vec{v}_r|} = \frac{-3\hat i - 4\hat j + 12\hat k}{\sqrt{(-3)^2 + (-4)^2 + 12^2}} = \frac{-3\hat i - 4\hat j + 12\hat k}{\sqrt{9 + 16 + 144}} = \frac{1}{13}(-3\hat i - 4\hat j + 12\hat k)$.

$A$ ray of light is incident on a convex mirror along a vector $3\hat i + 4\hat j + 12\hat k$. The normal to the convex mirror at the point of incidence is along $3\hat i + 4\hat j$. The unit vector along the reflected ray is:

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