Question

According to diagram, fish see two image of observer first through refraction only and other due to refraction and then reflection. Distance of the images from fish are $S_1$ and $S_2$ respectively observer can also see the two images of fish, first due to refraction only and other due to reflection and then refraction. Distance of them image from observer are $S_3$ and $S_4$ respectively :-

	COLUMN-I		COLUMN-II
(A)	$S_1$	(P)	$H\left[1+\frac{1}{2\mu }\right]$
(B)	$S_2$	(Q)	$H\left[\mu +\frac{1}{2}\right]$
(C)	$S_3$	(R)	$H\left[1+\frac{3}{2}\mu \right]$
(D)	$S_4$	(S)	$H\left[\mu +\frac{3}{2}\right]$

• A. $S_1\to s,S_2\to q,S_3\to r,S_4\to p$
• B. $S_1\to q,S_2\to s,S_3\to p,S_4\to r$
• C. $S_1\to q,S_2\to s,S_3\to r,S_4\to p$
• D. $S_1\to s,S_2\to q,S_3\to p,S_4\to r$

Answer 1

Answer: (B)

Due to refraction $1^{\text{st }}$ image (w.r.t. fish)
$S_1=\frac{H}{2}+\mu H$
due to reflection $S_2=\frac{H}{2}+(H+\mu H)$
w.r.t. (observer)
$S_3=\frac{H}{2\mu }+H;S_4=H+\mu H+\frac{H}{2}\mu$