Problem 6.98:A light ray is incident from air into a glass slab as shown in the diagram. Calculate the time spend by light ray inside glass. Refractive index of glass is $\sqrt{3}$.


Problem 6.98

Here the ray is making angle of $30^o$ with the plane of the slab. Angle of incidence is always measured with respect to the normal at the point of incidence.

So the angle of incidence, $i\:=\:60^o$.

The light ray suffers refraction and bends as shown in the next diagram. The ray travels inside glass along $PR$ and then emerges out.

Problem 6.98 B

Let us apply Snell’s law at the point of incidence, $P$.

We get $\frac{sin\: i}{sin\: r}\:=\:n$.

$\frac{sin \:60}{sin \:r}\:=\:\sqrt{3}$.

From this, $sin\:r\:=\:\frac {1}{2}$ .

Therefore angle of refraction $r\:=\:30^o$.

In Triangle $PQR$, $cos\:r\:=\:\frac{PQ}{PR}$

Or, $PR\:=\:\frac{PQ}{cos\:r}$

But $PQ\:=\:5\:cm$.

Then, $PR\:=\:\frac{5}{cos\:30}\:=\::\frac{5}{\sqrt{3}/2}$.

Or, $PR\:=\:\frac{10}{\sqrt{3}}\:cm\:=\:\frac{10}{\sqrt{3}}\:\times 10^{-2}\:m$.

Speed of light in glass, $v \:=\:\frac{c}{n}$ where $c$ is the speed of light in vacuum which is $3\times 10^8\:ms^{-1}$ and $n$ is the refractive index.

Then  $v \:=\:\frac{3\times 10^8}{\sqrt{3}}\:=\:\sqrt{3}\times 10^8\:ms^{-1}$.

Time  spent by light in glass slab, $t\:=\:\frac{PR}{v}\:=\:\frac{\frac{10}{\sqrt{3}}\:\times 10^{-2}}{\sqrt{3}\times 10^8}$,.

This will be approximately $3.33\times 10^{-10}$ second.