Question

Physics Question, No Spamming--->Reported
Refer to the attachment below, thanks :D

Answer 1

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Refer the attachment !!

First draw a normal for incident ray AB and reflected ray BC.

∠PBO = 90°

∠PBA = 38°

∴ ∠ABO = ∠PBO - ∠PBA

⇰ 90° - 38°

⇰ 52°

According to laws of reflection of light,

∠i = ∠r

∴ ∠ABO = ∠OBC = 52°

Now,

∠QBO = 90°

∠ OBC = 52°

∴ ∠QBC = ∠QBO - ∠OBC

⇰90° - 52°

⇰38°

In triangle BQC,

∠QBC = 38°

∠BQC = 90°

∴ ∠QCB = 180° - (∠QBC + ∠BQC)

⇰ 180° - (38° + 90°)

⇰ 180° - 128°

⇰ 52°

Now,

∠QCP = 90°

∠QCB = 52°

∴ ∠BCP = ∠QCP - ∠QCB

⇰ 90° - 52°

⇰ 38°

We know that,

∠i = ∠r

∠BCP = ∠DCP

38° = rf

∴ rf = 38° (ans)