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Question: What should be the angle between two plane mirrors, so that whatever be the angle of incident, the i...

What should be the angle between two plane mirrors, so that whatever be the angle of incident, the incident ray and the reflected ray from the two mirrors be parallel to each other?
(A)60\left( A \right)\,60^\circ
(B)90\left( B \right)\,90^\circ
(C)120\left( C \right)\,120^\circ
(D)45\left( D \right)\,45^\circ

Explanation

Solution

First we should draw a diagram of two plane mirrors placed at an angle to each other. The diagram shows a ray incident on the second mirror after reflecting from the first mirror then after from that second mirror it reflects in such a way that the reflected ray is parallel to that of the incident ray .

Complete step by step solution:

Here A and C are two mirrors where F and G are incident and reflected rays respectively. We need to find the DBE\angle DBE.
Let DH bisects the FDE\angle FDE into two equals part let say it as θ1{\theta _1} (HDE=FDH)(\angle HDE = \angle FDH)
Similarly EH bisects the DEG\angle DEG into two equal angle let say it as θ2(DEH=HEG){\theta _2}\left( {\angle DEH = \angle HEG} \right)
Now we can say that
FDH=HDE=θ1\angle FDH = \angle HDE = {\theta _1}
DEH=HEG=θ2\angle DEH = \angle HEG = {\theta _2}
According to the question ray F and G are parallel to each other
Hence
FDE+DEG=180\angle FDE + \angle DEG = 180^\circ
2θ1+2θ2=180\Rightarrow 2{\theta _1} + 2{\theta _2} = 180^\circ
θ1+θ2=90(1802)\Rightarrow {\theta _1} + {\theta _2} = 90^\circ \cdot \cdot \cdot \cdot \left( {\dfrac{{180^\circ }}{2}} \right)
We know the sum of all the angles in a triangle =180= 180^\circ.
Hence in DEH\vartriangle DEH ,
HDB+DEG+EHD=180\angle HDB + \angle DEG + \angle EHD = 180^\circ
θ1+θ2+EHD=180\Rightarrow {\theta _1} + {\theta _2} + \angle EHD = 180^\circ
We know that θ1+θ2=90{\theta _1} + {\theta _2} = 90^\circ
90+EHD=180\Rightarrow 90^\circ + \angle EHD = 180^\circ
EHD=90\Rightarrow \angle EHD = 90^\circ
As we can see from the diagram that DHBCDHBCis a rhombus ,which says that the opposite angles are equal to each other.
That’s why
EHD=DBE=90(Ans)\angle EHD = \angle DBE = 90^\circ (Ans)
Therefore, the correct option is (B)\left( B \right) .

Note: If the correct diagram is not made then it becomes difficult to solve this kind of problem and while making the diagram make sure that the rays are exactly parallel to each other to get an accurate answer to this kind problem.