A ray of light passes through a medium whose refractive index varies with distance as $$n=n01+x2a$$.ray enters the medium parallel to $$x-axis$$, what w be the time taken for ray to travel between $$x=0$$ an $$x=a$$ ?

Since, we know that μ$$=cv$$⇒ $$v=c$$μSo, if v be the speed at a distance x from $$y-axis$$, $$thenv=cn01+x2asince$$, $$v=dxdt$$, so we $$havedxdt=cn01+x2a$$ ⇒∫$$01dt=n0c$$∫$$0a1+x2adx$$ $$t=5n0a4c$$

Questions

A ray of light passes through a medium whose refractive index varies with distance as $n = n_{0} (1 + \frac{x}{2 a})$ .ray enters the medium parallel to x-axis, what w be the time taken for ray to travel between x=0 an x=a ?

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t = \frac{5 n_{0} a}{3 c}

t = \frac{5 n_{0} a}{7 c}

t = \frac{6 n_{0} a}{4 c}

t = \frac{5 n_{0} a}{4 c}

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detailed solution

Correct option is D

Since, we know that $μ = \frac{c}{v}$

$\Rightarrow v = \frac{c}{μ}$

So, if v be the speed at a distance x from y-axis, then

$v = \frac{c}{n_{0} (1 + \frac{x}{2 a})}$

since, $v = \frac{d x}{d t}$ , so we have

$\frac{d x}{d t} = \frac{c}{n_{0} (1 + \frac{x}{2 a})} \Rightarrow \int_{0}^{1} d t = \frac{n_{0}}{c} \int_{0}^{a} (1 + \frac{x}{2 a}) d x t = \frac{5 n_{0} a}{4 c}$