Atank whose bottom is a mirror is filled with water to a depth of 20.6 cm . a small fish floats motionless 7.80 cm under the surface of the water. part a
what is the apparent depth of the fish when viewed at normal incidence to the water?
express your answer in centimeters. use 1.33 for the index of refraction of water.
part b
what is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?
express your answer in centimeters. use 1.33 for the index of refraction of water.

Explanation:

A )  Index of refraction = real depth / apparent depth

real depth = 7.8 cm  ; Index of refraction = 1.33

1.33 = 7.8 /apparent depth

apparent depth = 7.8 / 1.33

= 5.86 cm

B )  

Distance of fish fro mirror = 20.6 - 7.8 = 12.8 cm

Image will be formed 12.8 cm behind the mirror.

Distance of image from the water surface = 20.6 +12.8

= 33.4 cm

Real depth of image = 33.4

Apparent depth of image  = 33.4 / 1.33

= 25.11 cm

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s