A bird is flying over a swimming pool at a height of `2m` from the water surface. If the bottom is perfectly plane reflecting surface and depth of swimming pool is `1m,` then the distance of final image of bird from the bird itself is `mu_w=4//3`A. `11/3 m`B. `23/3 m`C. `11/4 m`D. `11/2 m`

A bird is flying over a swimming pool at a height of `2m` from the wat

1.	A bird is flying over a swimming pool at a height of `2m` from the water surface. If the bottom is perfectly plane reflecting surface and depth of swimming pool is `1m,` then the distance of final image of bird from the bird itself is `mu_w=4//3`A. `11/3 m`B. `23/3 m`C. `11/4 m`D. `11/2 m`
Answer» Correct Answer - D `h_(app)=muh=4/3xx2=8/3m` `Distance from mirror=1+h_(app)=11/3m.` So, mirror will make image at same distance (`=11//3m` from itself). Now in third refraciton, depth of second image, `d=11/3+1=14/3 m` `d_(app)=d/mu=14/3xx3/4=7/2m` The desired distance is therefore, `(d_(app)+h)` or `(7/2+2)m or 11/2m`

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A bird is flying over a swimming pool at a height of `2m` from the water surface. If the bottom is perfectly plane reflecting surface and depth of swimming pool is `1m,` then the distance of final image of bird from the bird itself is `mu_w=4//3`A. `11/3 m`B. `23/3 m`C. `11/4 m`D. `11/2 m`

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