Q.A monochromatic ray of light strike the water (n = 4/3) surface in a cylindrical vessel at angle of incidence 53°. Depth of water is 36 cm. After striking the water surface, how long will the light take to reach the bottom of the vessel? [Angles of the most popular Pythagorean triangle of sides in the ratio 3:4:5 are nearly 37°, 53° and 90°]

Using real depth h and apparent depth h/n (n = 4/3, water's standard refractive index), the apparent rise h(1 - 1/n) equals 1 m when h = 4 m, so the tank should be filled to a 4 m water level. Note: this question is on Ray Optics (apparent depth due to refraction), a topic not present in the 14-chapter menu supplied for this course, so it has been flagged out of syllabus for this course's chapter list, even though the physics itself is answerable.

Let the dead snake lie at the very bottom of the tank, and let h = the height (depth) of water poured in above it (so the snake is at a real depth h below the water surface).

When viewed (roughly vertically) from outside/above through the water surface, the apparent depth of the snake below the surface is:

apparent depth = h / n

where n is the refractive index of water (this value is not stated in the question, so the standard assumed value n = 4/3 is used, as is conventional in NCERT-level problems).

The apparent RISE in the snake's position (how much higher it appears to be, compared to its true position) is the difference between the real depth and the apparent depth:

apparent rise = h - h/n = h (1 - 1/n) = h (n-1)/n

The apparent rise needed is such that the snake's APPARENT position is 1 m above the actual bottom of the tank, i.e. apparent rise = 1 m:

h (n-1)/n = 1

With n = 4/3: (n-1)/n = (1/3)/(4/3) = 1/4

h / 4 = 1

h = 4 m

So pouring water to a level (height) of 4 m (leaving the top 1 m of the 5 m tall tank as air, since the tank height is 5 m) makes the snake's apparent position appear raised by 1 m above the real bottom, making it visible to the observer.