An insect starts moving up in a liquid from p | vedclass

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Question

(D) The apparent depth $h$ of an object at a depth $y$ in a medium with variable refractive index $\mu(y)$ is given by the integral $h = \int_{0}^{y}

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) The apparent depth $h$ of an object at a depth $y$ in a medium with variable refractive index $\mu(y)$ is given by the integral $h = \int_{0}^{y} \frac{dy'}{\mu(y')}$.Given $\mu = \mu_0(1 + ay)$,the apparent depth $h$ is:$h = \int_{0}^{y} \frac{dy'}{\mu_0(1 + ay')} = \frac{1}{\mu_0 a} [\ln(1 + ay')]_{0}^{y} = \frac{\ln(1 + ay)}{\mu_0 a}$.The apparent speed $v_{app}$ is the rate of change of apparent depth with respect to time:$v_{app} = \frac{dh}{dt} = \frac{d}{dt} \left( \frac{\ln(1 + ay)}{\mu_0 a} \right)$.Using the chain rule,$\frac{dh}{dt} = \frac{1}{\mu_0 a} \cdot \frac{1}{1 + ay} \cdot \frac{d(1 + ay)}{dt} = \frac{1}{\mu_0 a} \cdot \frac{1}{1 + ay} \cdot (a \frac{dy}{dt})$.Since the speed of the insect is $u = -\frac{dy}{dt}$ (as $y$ decreases as it moves up),we have $\frac{dy}{dt} = -u$.Thus,the magnitude of the apparent speed is $v_{app} = \frac{u}{\mu_0(1 + ay)}$.Comparing this with the given options,none of the expressions match the derived result exactly as a constant,as it depends on the position $y$. Therefore,the correct choice is 'None of these'.

An insect starts moving up in a liquid from point $O$ with a variable refractive index $\mu = \mu_0(1 + ay)$,where $y$ is the depth of the liquid from the surface. If $u$ is the speed of the insect,what is its apparent speed to the observer $E$?

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