Question

A fish is inside a water container, whose bottom is polished to make it a plane mirror. The fish is observing the image of bird in the mirror. The distance between image of bird and fish in the figure given below is [Take refractive index of water as ) TH 102 d (1) d + uH (2) زانا + 1+uH 3d +H NIW d 3 2 (3)​

Answer 1

Fish observing eye:

(i) Direct observation:

H 1 = 2H+μH 1 =H( 21 +μ)

(ii) Fish observing image of eye by mirror.

Hence, distance of the eye image from fish,H 2 =μH+H+ 2H

H 2 =H( 23+μ)

Eye observing fish:

(i) Direct observation H 1=H+ 2μH =H(1+ 2μ1 )

(ii) Eye observing image of the fish H 2′ =H+ μH2μH =H(1+ μ2 )H 2′ =H+ 2μ3 .

Answer 2

Answer:

The correct answer is $\mathrm{H}+2 \mu _{3}$.

Explanation:

Step 1

Fish observing eye:

(i) Direct observation:

$\mathrm{H} _{1} =2 \mathrm{H}+\mu \mathrm{H} _{1} =\mathrm{H}(21+\mu)$

(ii) Fish observes an image of the eye in the mirror.

Hence, the distance of the eye image from fish, $\mathrm{H} _{2} =\mu \mathrm{H}+\mathrm{H}+2 \mathrm{H}$

$\mathrm{H} _{2} =\mathrm{H}(23+\mu)$

Step 2

Eye observing fish:

(i) Direct observation

$\mathrm{H} \mathrm{} _{1} =\mathrm{H}+2 \mu \mathrm{H}=\mathrm{H}(1+2 \mu _{1} )$

(ii) Eye observing an image of the fish $\mathrm{H} _{2} ^{\prime}=\mathrm{H}+\mu \mathrm{H} _{2} \mu \mathrm{H}$

$=\mathrm{H}(1+\mu _{2} ) \mathrm{H} _{2} ^{\prime}=\mathrm{H}+2 \mu _{3}$

Therefore, the correct answer is $\mathrm{H}+2 \mu _{3}$.

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