m and n are plane mirrors perpendicular to each other. Prove that incident ray CA (to mirror n ) is parallel to reflected ray (to mirror m ) BD. 

Respected Sirs,

Please answer my this question as soon as possible. I tried to insert the figure from ms-paint file but was unable to do so. But I believe that the question is self descriptive.

Dear student

Given: Mirrors m and n are prependicular to each other.The incident ray CA is reflected along AB and again re-reflected along BD.To prove: CABDConstruction: Draw normals AM and BN.Proof: Since AM is a normal to the mirror n and CA is incident ray and AB is reflectedray, therefore,2=1    [by laws of reflection]Similarly, incident ray AB , reflected ray BD and BN is the normal at the point ofincidence.3=4      [By laws of relection]Since the mirror mn and MAn.Mirror mMA and AB is a transversal.1=5       ...(1)[Alternate interior angles]Also 5+3=90°    [BNMorror m]1+3=90°    [5=1]21+23=2×90°CAB+ABD=180°CAB and ABD are consecutive interior angles on the same side of transversal AB.CABD.

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Let us draw BM ⊥ PQ and CN ⊥ RS.

As PQ || RS,

Therefore, BM || CN

Thus, BM and CN are two parallel lines and a transversal line BC cuts them at B and C respectively.

∴∠2 = ∠3 (Alternate interior angles)

However, ∠1 = ∠2 and ∠3 = ∠4 (By laws of reflection)

∴ ∠1 = ∠2 = ∠3 = ∠4

Also, ∠1 + ∠2 = ∠3 + ∠4


However, these are alternate interior angles.

∴ AB || CD

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I am not able to view the image?!?
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This is not the answer to this question. We should prove that the angle of incidence is equal to the angle of reflection.
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bhen ncert mein hai ye qs....kahaan khoye hue ho ??  

XD  rofl !!
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