Can you help him out?

One way to solve this problem is by kinematics (as done by abhineet here). Now let us try another method i.e by geometrical optics.

Let us assume light ray travel as shown in figure.

image

Region marked 1 is where light travels n times slower.(refractive index = $\eta$ ) Region marked 2 is where light travels with normal speed in air.(refractive index =1)

$\frac { v }{ \eta } =\frac { v }{ n }$

$\eta$ = $n$

So by snell's law

Sin(a)= $\frac { 1 }{ \eta }$

Hence tan (a)= $\frac { 1 }{ \sqrt { { n }^{ 2 }-1 } }$

$l$ tan(a)= $\frac { l }{ \sqrt { { n }^{ 2 }-1 } }$ .

put l=500 and n=3

Let the speed of the car on the highway be $'v'$ . So, it's speed in the field is $\frac { v }{ 3 }$ . Let the total distance from $A\quad to\quad D$ is $'S'$ and let the distance of the diverting point(Let it be $'M'$ ) from $D = 'x'$ .

Time taken by the car in moving from $A\quad to\quad M = \frac { Distance }{ Speed } =\frac { S-x }{ v }$

in Triangle BND, $\quad { BN }^{ 2 }={ x }^{ 2 }+{ l }^{ 2 }\\ BN=\sqrt { { x }^{ 2 }+{ l }^{ 2 } }$

Time taken in moving from $M\quad to\quad B = \frac { Distance }{ Speed } =\frac { \sqrt { { x }^{ 2 }+{ l }^{ 2 } } }{ \frac { v }{ 3 } } =\frac { 3\sqrt { { x }^{ 2 }+{ l }^{ 2 } } }{ v }$

Total time taken (Let it be $'t'$ ) = $\frac { S }{ v } -\frac { x }{ v } +\frac { 3\sqrt { { x }^{ 2 }+{ l }^{ 2 } } }{ v }$ .

Since, we need to find the minimum time, we can differentiate it with respect to $'x'$ and equate it to $0$ .

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \frac { dt }{ dx } =0\\ \Rightarrow \frac { d }{ dx } (\frac { S }{ v } -\frac { x }{ v } +\frac { 3\sqrt { { x }^{ 2 }+{ l }^{ 2 } } }{ v } )=0\\ \\ \Rightarrow \frac { d }{ dx } (\frac { -x }{ v } )+\frac { 3 }{ v } \frac { d }{ dx } \sqrt { { x }^{ 2 }+{ l }^{ 2 } } =0\\ \\ \Rightarrow \frac { -1 }{ v } +\frac { 3 }{ 2v\sqrt { { x }^{ 2 }+{ l }^{ 2 } } } (\frac { d }{ dx } { x }^{ 2 }+{ l }^{ 2 })=0\\ \\ \Rightarrow \frac { 3\quad \times \quad 2x }{ 2v\sqrt { { x }^{ 2 }+{ l }^{ 2 } } } \quad =\quad \frac { 1 }{ v } \\ \\ \Rightarrow \quad 3x=\sqrt { { x }^{ 2 }+{ l }^{ 2 } } \\ \Rightarrow \quad 9{ x }^{ 2 }\quad =\quad { x }^{ 2 }+{ l }^{ 2 }\\ \Rightarrow \quad 8{ x }^{ 2 }\quad =\quad { l }^{ 2 }\\ \Rightarrow \quad x\quad =\quad \frac { l }{ 2\sqrt { 2 } } \\ \\ Putting\quad l=500m,\quad We\quad get\\ x\quad =\quad 176.776m$

CHEERS!!:)

In general using derivative bashing we can say that $D_{min} = \frac{L} {\sqrt(\alpha^2-1)}$

Its some starting Irodov problem dont exactly recall did it sometime back last month.