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. $S_{1} \rightarrow s,S_{2} \rightarrow q,S_{3} \rightarrow r,S_{4} \rightarrow p$
• B. $S_{1} \rightarrow q,S_{2} \rightarrow s,S_{3} \rightarrow p,S_{4} \rightarrow r$
• C. $S_{1} \rightarrow q,S_{2} \rightarrow s,S_{3} \rightarrow r,S_{4} \rightarrow p$
• D. $S_{1} \rightarrow s,S_{2} \rightarrow q,S_{3} \rightarrow p,S_{4} \rightarrow r$

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 $S_{2}=\frac{H}{2}+\left(\right.H+\mu H\left.\right)$
w.r.t. (observer)
$S_{3}=\frac{H}{2 \mu }+H;S_{4}=H+\mu H+\frac{H}{2}\mu $