A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank. a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u  = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance  = -  .465 m

1 / v + 1.333 / .465  = (1 -1.333 )/1.95

1 / v + 2.8667  = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u ,  μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

explanation:

an enlarged virtual image is the answer

1 a

2 b

3 b

4 d

5 d

6 d

explanation: