Question
Physics Question on Ray optics and optical instruments
A ray is incident on a plane surface. If i^+j^−k^ represents a vector along the direction of incident ray.i^+j^ is a vector along normal on incident point in the plane of incident and reflected ray, then vector along the direction of reflected ray is
−191(−3i^+3j^+k^)
191(3i^+3j^−k^)
−31(i^+j^+k^)
k^
−31(i^+j^+k^)
Solution
According to law of reflection in vector form,
n2^=n1^−2(n1^.n^)n^
Here, n1^ = the unit vector along incident ray
=3i^+j^−k^
n^ = unit vector along normal on incident point
=2i^+j^
n2^ =unit vector along the direction of reflected ray
Using the formula, we get
\hat{n_2} = \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - 2 \bigg\\{ \bigg( \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}\bigg) . \bigg( \frac{\hat{i} + \hat{j}}{\sqrt{2}} \bigg). \bigg( \frac{\hat{i} + \hat{j} }{\sqrt{2}} \bigg) \bigg\\}
= \frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}} - 2 \bigg\\{ \frac{1}{\sqrt{6}} + \frac{1}{\sqrt{6}} \bigg\\} \frac{\hat{i} + \hat{j}}{\sqrt{2}}
=3i^+j^−k^−64(2i^+j^)
=3i^+j^−k^−124i^−124j^
=232i^+2j^−2k^−4i^−4j^ (∵12=23)
=23−2i^−2j^−2k^=3−i^−j^−k^
=−31(i^+j^+k^)