If hat{n}_1, hat{n}_2 and hat{t} represent un | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $\hat{t}$ be the unit vector along the normal to the surface.Let $\hat{n}_1$ be the unit vector along the incident ray and $\hat{n}_2$ be the

Vedclass · Accepted Answer

$\hat{n}_2=\hat{n}_1-2(\hat{n}_1 \cdot \hat{t}) \hat{t}$

Vedclass · Answer

(A) Let $\hat{t}$ be the unit vector along the normal to the surface.Let $\hat{n}_1$ be the unit vector along the incident ray and $\hat{n}_2$ be the unit vector along the reflected ray.From the geometry of reflection, the angle of incidence equals the angle of reflection, $	heta$.We can resolve $\hat{n}_1$ and $\hat{n}_2$ into components parallel and perpendicular to the normal $\hat{t}$.$\hat{n}_1 = -\cos 	heta \hat{t} + \sin 	heta \hat{u}$, where $\hat{u}$ is a unit vector parallel to the surface.$\hat{n}_2 = \cos 	heta \hat{t} + \sin 	heta \hat{u}$.Subtracting the two equations: $\hat{n}_2 - \hat{n}_1 = 2 \cos 	heta \hat{t}$.Since $\hat{n}_1 \cdot \hat{t} = \cos(180^\circ - 	heta) = -\cos 	heta$, we have $\cos 	heta = -(\hat{n}_1 \cdot \hat{t})$.Substituting this into the subtraction equation: $\hat{n}_2 - \hat{n}_1 = -2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.Therefore, $\hat{n}_2 = \hat{n}_1 - 2(\hat{n}_1 \cdot \hat{t}) \hat{t}$.

If $\hat{n}_1, \hat{n}_2$ and $\hat{t}$ represent unit vectors along the incident ray, reflected ray, and normal to the surface respectively, then:

Similar Questions

$A$ mirror is inclined at an angle of $\theta$ with the horizontal. If a ray of light is incident at an angle $\theta$ with the mirror,then the reflected ray makes the following angle with the horizontal:

Which of the following sets of letters does not show lateral inversion?

The figure shows two rays $A$ and $B$ being reflected by a mirror and emerging as $A'$ and $B'$. The mirror

$A$ boy of height $1.5\, m$ with his eye level at $1.4\, m$ stands before a plane mirror of length $0.75\, m$ fixed on the wall. The height of the lower edge of the mirror above the floor is $0.8\, m$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module