A) Moving faster than its speed and also away from the real distance
B) Moving faster than its real speed and nearer than its real distance
C) Moving slower than its real speed and also nearer than its real distance
D) Moving slower than its real speed and away from the real distance
Correct Answer: A
Solution :
[a] \[\frac{{{\mu }_{1}}}{-u}+\frac{{{\mu }_{2}}}{v}=\frac{{{\mu }_{1}}-{{\mu }_{2}}}{R}\] For a plane surface, \[R=\infty \] \[\therefore \frac{{{\mu }_{1}}}{-u}+\frac{{{\mu }_{2}}}{v}=0\] or \[\frac{{{\mu }_{2}}}{v}=\frac{{{\mu }_{1}}}{u}\] or \[\frac{\mu }{v}=\frac{1}{u}\] Or \[v=\mu u\] Clearly, to the fish, the bird appears farther than its actual distance. Again, \[\frac{dv}{dt}=\mu \frac{du}{dt}\] Or. Apparent speed of bird = refractive index \[\times \]Actual speed of bird
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