Question

A person lives in a high-rise building on die 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 fight bulbs at height $Y$ on the tower. The values of $X$ and $Y$ are respectively close to (refractive index of water $\simeq \frac{4}{3}$ )

• A. 25 m, 10 m
• B. 13 m, 27 m
• C. 22 m, 13 m
• D. 17 m, 20 m

Answer 1

Answer: (B)

$i =$ Brewster's angle
$=\tan ^{-1}(\mu)$
$=\tan ^{-1}\left(\frac{4}{3}\right)$
$i =53^{\circ}$
$\therefore \tan 37^{\circ}=10 / X$
$\frac{3}{4}=\frac{10}{ X }$
$\Rightarrow \frac{40}{3}=13.3\, m$
Again, $\tan 37^{\circ}=\frac{ Y }{50- X }$
$\frac{3}{4}=\frac{ Y }{50- X }$
$\Rightarrow 3\left(50-\frac{40}{3}\right)=4 Y$
$\therefore 110=4 \,Y =27.5\, m$