A spherical fish bowl of radius R is placed in front of a plane vertical mirror (M). The thickness of the wall of the fish bowl is very thin. The centre (C) of the spherical bowl is at a distance of 3R from the plane mirror. The bowl is filled with water and contains a fish (F). Fish (F) is at a distance of R from the centre of the spherical bowl as shown in the figure. Refractive index of water is . Two surfaces are indicated in the bowl as first surface (1) and second surface (2)Column-I (Optical Event)Column-II (Nature of image)(A) Refraction at first surface(P) Virtual (B) Refraction at second surface after reflection from mirror(Q) Real(C) Refraction at first surface after reflection from mirror and refraction from second surface.(R) Magnified(S) Diminished

Engineering

Physics

Curved Surface Refraction

Question

A spherical fish bowl of radius R is placed in front of a plane vertical mirror (M). The thickness of the wall of the fish bowl is very thin. The centre (C) of the spherical bowl is at a distance of 3R from the plane mirror. The bowl is filled with water and contains a fish (F). Fish (F) is at a distance of R from the centre of the spherical bowl as shown in the figure. Refractive index of water is $\frac{4}{3}$ . Two surfaces are indicated in the bowl as first surface (1) and second surface (2)

Column-I (Optical Event)	Column-II (Nature of image)
(A) Refraction at first surface	(P) Virtual
(B) Refraction at second surface after reflection from mirror	(Q) Real
(C) Refraction at first surface after reflection from mirror and refraction from second surface.	(R) Magnified
	(S) Diminished

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$\frac{µ_{r}}{v} - \frac{µ_{i}}{u} = \frac{µ_{r} - µ_{i}}{R}$

$\frac{1}{v} - \frac{4 / 3}{(- 2 R)} = \frac{1 - 4 / 3}{(- R)}$ $\frac{1}{v} + \frac{2}{3 R} = - \frac{1}{3 R}$

v = –3R (virtual)
$m = \frac{µ_{i}}{µ_{r}} = \frac{4 / 3}{(1)} \frac{(- 3 R)}{(- 2 R)}$
m = 2 (magnified)
Reflection from mirror & refraction from second surface

$\frac{4 / 3}{v} - \frac{1}{(- 4 R)} = \frac{4 / 3 - 1}{R}$
v = 16R (Real)
$m = \frac{µ_{i}}{µ_{r}} \frac{v}{u} = \frac{1}{4 / 3} \frac{16 R}{(- 4 R)} \Rightarrow$ m = –3 (magnified)
Refraction a first surface after reflection from mirror and refraction at second surface.

$\frac{µ_{r}}{v} - \frac{µ_{i}}{u} = \frac{µ_{r} - µ_{i}}{R}$

$\frac{1}{v} - \frac{4 / 3}{14 R} = \frac{1 - 4 / 3}{- R}$
v = $\frac{7 R}{3}$ (Real)
m = $\frac{µ_{i}}{µ_{r}} \frac{v}{u}$
$m = \frac{4}{3} \times \frac{7 / 3 R}{+ 14 R} = \frac{4}{3} \times \frac{7}{3 \times 14} \Rightarrow m = \frac{2}{9}$ (diminished)