KCET · Physics · Ray Optics
A fish in water (refractive index \(n\) ) looks at a bird vertically above in the air. If \(y\) is the height of the bird and \(x\) is the depth of the fish from the surface, then the distance of the bird as estimated by the fish is
- A \(x+y\left(1-\frac{1}{n}\right)\)
- B \(x+n y\)
- C \(x+y\left(1+\frac{1}{n}\right)\)
- D \(y+x\left(1-\frac{1}{n}\right)\)
Answer & Solution
Correct Answer
(B) \(x+n y\)
Step-by-step Solution
Detailed explanation
When object is in rarer medium and observer is in denser medium
Normal shift, \(\quad \mathrm{d}=(\mathrm{n}-1) \mathrm{h}\)
where \(h=\) real depth
Here, \(\quad h=y\)
Now, apparent depth or the apparent height of the bird from the surface of the water \(=\mathrm{y}+(\mathrm{n}-1) \mathrm{y}=\mathrm{ny}\).
The total distance of the bird as estimated by fish is \(x+n y\).
Normal shift, \(\quad \mathrm{d}=(\mathrm{n}-1) \mathrm{h}\)
where \(h=\) real depth
Here, \(\quad h=y\)
Now, apparent depth or the apparent height of the bird from the surface of the water \(=\mathrm{y}+(\mathrm{n}-1) \mathrm{y}=\mathrm{ny}\).
The total distance of the bird as estimated by fish is \(x+n y\).
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