question_answer

                A person lives in a high-rise building on the bank of a river 50 m wide. Across the river is a well lit tower of height 40 m. When the person, who is at a height of 10 m, looks through a polarizer at an appropriate angle at light of the tower reflecting from the river surface, he notes that intensity of light coming from distance X from his building is the least and this corresponds to the light coming from light bulbs at height ?Y? on the tower. The values of X and Y are respectively close to (refractive index of water\[\simeq \frac{4}{3}\])                   JEE Main Online Paper (Held On 09 April 2013)

A)                 25 m, 10 m                                         

B)                 13 m, 27 m                                         

C)                 22 m, 13 m                         

D)                 17 m, 20 m 

Correct Answer: C

Solution :

                                               It is the case of Booster angle where reflected and refracted ray both become polarized. So,          \[{{\mu }_{1}}=\tan {{i}_{B}}\]                 \[\frac{4}{3}=\tan {{i}_{B}}\]                 From triangle ABC,                 \[\tan (90-{{i}_{B}})=\frac{BC}{AB}\]                 \[\cot \,{{i}_{B}}=\frac{10}{x}\]                 \[\frac{3}{4}=\frac{10}{x}\]                 \[x=\frac{40}{3}=13\,m\]                 From triangle AEF,                 \[\tan (90-{{i}_{B}})=\frac{y}{50-x}\]                 \[\cot {{i}_{B}}=\frac{y}{50-13}\]                 \[\frac{3}{4}=\frac{y}{37}\]                 \[y=\frac{3\times 37}{4}\]                 \[y=27\,m\]