A jellyfish is floating in a water-filled aquarium 1.38 m behind a flat pane of

Question

A jellyfish is floating in a water-filled aquarium 1.38 m behind a flat pane of glass 4.00 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.)

Explanation / Answer

n1 = 1.5; n2 = 1.33

the appearnce of width of fish remains same:

sin = tan = x/l

Applying Snell law:

n1sin1 = n2sin2

ni*x / li = nf*x / lf

lf = li/ni

d = l2/n2 +l1/n1

= 1.38/1.33 + 0.04/1.5

= 1.064 m

The fish apperas to be closer by:

= 1.38+0.04-1.064

=0.356 m

(b)

When the thickness is negligible the image of fish will be formed as per the law of refraction:

depth' = actual depthx n(observer) / n (object)

   = 1.38 x 1 / 1.33

= 1.0375 m

This is the distance of fish image from observer

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