Question

in fig6.33,PQ and rs are two mirrors placed parallel to each other an incidences ray move along the path bc and strikes the mirror rs at c and again reflects back along CD prove that ab||cd answer cf=rs​

Answer 1

Given :

• PQ and RS are two mirrors parallel to each other.

To Prove :

• AB // CD

Solution :

We will prove ∠ABC = ∠BCD :

Construct a line BM normal to PQ and CN normal to RS.

By law of reflection , angle of reflection = angle of incidence.

$\longmapsto\tt{\angle{1}=\angle{2}----(1)}$

$\longmapsto\tt{\angle{3}=\angle{4}----(2)}$

$\longmapsto\tt{\angle{2}=\angle{3}----(3)}$

Now , BM // CN :

$\longmapsto\tt{\angle{1}=\angle{2}=\angle{3}=\angle{4}}$

$\longmapsto\tt{\angle{ABC}=\angle{3}+\angle{4}}$

$\longmapsto\tt\bold{\angle{ABC}=\angle{BCD}}$

Since Alternate Angles are equal .So , the lines are parallel..

Answer 2

Question:-

In the figure,PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB||CD.

Answer:-

At surface B, ∠1=∠2(property of reflection)

At surface C,∠3=∠4(property of reflection)

As the sum of the angles are equal AB is parallel to CD,

∠2=∠3(Alternate interior angles)

∠1+∠2=∠3+∠4

Alternate interior angles are equal.

∴AB||CD.