Question

A vessel having perfectly reflecting plane bottom is filled with water \( (\mu= \) \( 4 / 3 \) ) to a depth \( d \). A point source of light is placed at a height \( h \) above the surface of water. Find the distance of final image from water surface.

(1) \( h+\frac{5}{2} d \)

(2) \( h-\frac{3}{2} d \)

(3) \( h+\frac{4}{3} d \)

(4) \( h+\frac{3}{2} d \)

Answer 1

The correct option is (4) \(h + \frac{3}{2} d\)

Answer 2

Correct option is: (4)  \(h + \frac{3}{2}d\)

As shown in Fig. water form the image of object O at  \(I_1\)  such that  \(OI_1 = y = d[1 - (\frac{1}{\mu})]\), so that the distance of image \(I_1\)  from water surface will be \(I_1 A = h - y = h - d [1 - (\frac{1}{\mu})]\) . Hence, the distance of this image \(I_1\) from mirror \(MM'\) , \(I_1 M = I_1A + AM\)

\(= [h - d( 1 - \frac{1}{- \mu})] + d = h + [\frac{d}{\mu}]\)

Now, image \(I_1\) will act as object for mirror \(MM'\). As a plane mirror forms image at same distance behind the mirror as the object is in front of is, the image of \(I_1\) formed  by the mirror \(MM'\) will be \(I_2\) such that \(I_1 M = I_2 M = h + (\frac{d}{\mu})\) .

Now, this image \(I_2\) will act as object for water again and water will produce image \(I_3\) such that \(I_2I_3 = y = d(1 - \frac{1}{\mu})\) . So, the distance of image \(I_3\) from the surface of water AC will be

\(AI_3 = AM + MI_2 = I_2 I_3 = d + [h + \frac{d}{\mu}] - d[1 - \frac{1}{\mu}]\)

\(AI_3 = h + 2 \frac{d}{\mu} = h + \frac{3}{2}d\)