Question

Please type out solution

. A jellyfish is floating in a water-filled aquarium 1.46 m behind a flat pane of glass 3.90 cm thick and having an index of refraction of 1.5. (The index of refraction of water is 1.33.)

(a) Where is the image of the jellyfish located? (Find the location of the final image, taking the inside and outside surfaces of the pane of glass as two refracting surfaces. Use a negative number to indicate the image is in the tank. Measure the distance from the air-glass interface. Enter your answer to three decimal places.)

(b) Repeat the problem when the glass is so thin that its thickness can be neglected. (Use a negative number to indicate the image is in the tank. Enter your answer to three decimal places.)

Answer 1

Given,Thickness, t = 3.9 cm = 0.039 m

The refractive index of a glass, n₁ = 1.5

The refractive index of water, n₂ = 1.33

D = 1.46 m

(a)

The jelly fish is located at:

The refractive index, n = D₀ / D_i

Where, n = n_{1 *} n₂

= 1.5 * 1.33 = 1.995

D₀ = D + t

= 1.46 + 0.039 = 1.499 m

As, n = D₀ / D_i

=>D_i =D₀ / n

D_i = 1.499 / 1.995

D_i = 0.7514 m = 75.14 cm

(b)

When t = 0, the thickness is neglected

n = D₀ / D_i

D₀= D = 1.46 m

n = n₂ = 1.33

=> D_i = D₀ / n

= 1.46 / 1.33

= 1.0977 m

= 109.77 cm