Reflection Refraction, Fish Mirror

[gohindha89]

A tank whose bottom is a mirror is filled with water to a depth of 61.0 cm. A small fish floats motionless 21.0 cm under the surface of the water.



(a) What is the apparent depth of the fish when viewed at normal incidence?
(b) What is the apparent depth of the image of the fish when viewed at normal incidence?


I looked in a lot of places, but the usual consensus is that the formula needed is : na/ds + nb/di = 0 ( ds = distance of object, di = distance of image) I don't know what else to do, but I tried 1/21 +1.33/ds = 0, or using nasin(θA) = nb sin(θB), and using that to get distances which again didn't work. I don't know how to comprehend a question like this. Please help.

 

[jbsteele]

(a) The apparent depth of the fish when viewed at normal incidence is 21.0 cm.(b) The apparent depth of the image of the fish when viewed at normal incidence is 42.0 cm. This can be determined using the formula n*d1 + n*d2 = 0, where n is the refractive index of water and d1 and d2 are the distances of the object and the image, respectively. In this case, n = 1.33, d1 = 21 cm, and d2 = 42 cm. Therefore, 1.33*21 + 1.33*42 = 0.