
From the above figure, i2=30∘,
Now,
∠OBA+∠ABR+∠RBP=180∘⇒2∠OBA+60∘=180∘⇒∠OBA=60∘
In triangle OAB,
∠OBA+∠AOB+∠OAB=180∘⇒60∘+75∘+∠OAB=180∘⇒∠OAB=45∘
And,
∠OAB+∠BAI+∠IAQ=180∘⇒2∠OAB+2i1=180∘⇒i1=45∘
Now the total deviation of the ray will be,
δ=360∘−2(i1+i2)⇒δ=360∘−2(30∘+45∘)δ=210∘