Wavelength of light in a dielectric material

[laurieke]

Homework Statement

In a physics lab, light with a wavelength λ0 travels in vacuum from a laser to a photocell in a time t1. When a slab of glass with a thickness d is placed in the light beam, with the beam incident along the normal to the parallel faces of the slab, it takes the light a time of t2 to travel from the laser to the photocell (again in vacuum). The speed of light in a vacuum is c.
9. What is the wavelength of the light in the glass? Express your answer in the given constants.

Homework Equations

labda = labda0 / n
n = c/v
v=x/t

The Attempt at a Solution

so i said that v = x/t equals d/(t2-t1)
n = c/v so together with the above equation n = c(t2-t1)/d
i filled this in at the first equation labda/ labda0 / n and with this i got to the answer:
labda = labda0*d/c(t2-t1)
the answer on the other hand is : labda0*d/c(t2-t1)+d , i am very close but i can not figure out where the +d comes from

 

[Phylosopher]

so i said that v = x/t equals d/(t2-t1)

Here you are wrong. Draw everything to visualize, and instead of starting with velocity. Can you first find what is $\Delta t=t_{2}-t_{1}$?

 

[laurieke]

i do not understand what you mean, i only have the information in the question so dt = t2 -t1. I do not have numbers

 

[Phylosopher]

i do not understand what you mean, i only have the information in the question so dt = t2 -t1. I do not have numbers

I know. I will explain more.

If the total distance from the laser to the photocell is $L$. This distance is fixed when you put the glass slab in the middle, right? This means, in case (1) when there is no slab and in case (2) when there is a slab, the total distance is $L$.

But, in case (1) the laser passes the whole distance $L$, with speed $c$ from the beginning to the end. While in case (2), it passes $L-d$ with speed $c$ "In vacuum"... then passes $d$ with speed $v$ "In the slab".

Now, using these information write down $\Delta t=t_{2}-t_{1}$

 

[laurieke]

so it passes L-d with speed c first so t1 is (L-d)/c and after this it passed d with speed v so t2 is d/v?

 

[Phylosopher]

so it passes L-d with speed c first so t1 is (L-d)/c and after this it passed d with speed v so t2 is d/v?

No.

$t_{1}$ is the time it takes light to travel $L$ distance, from the laser to the photocell without the presence of glass.
$t_{2}$ is the time it take light to travel $L$ distance from the laser to the photocell with the presence of glass along the way.

For $t_{1}$, light travel all the way with speed $c$. Thus, $t_{1}=\frac{distance}{speed}=\frac{L}{c}$.

For $t_{2}$, light travel distance $L-d$ with speed $c$ and distance $d$ with speed $v$. Thus, $t_{1}=\frac{distance}{speed}=\frac{L-d}{c}+\frac{d}{v}$

 

[rude man]

Oops, had to delete a wrong answer.

There is a typo in post 6: should be t2 instead of t1 for last expression. But otherwise OK.
So how about using those equations for t1 and t2 to solve for v/c?

 

[Phylosopher]

$t_{1}=\frac{distance}{speed}=\frac{L-d}{c}+\frac{d}{v}$

Correction: $t_{2}=\frac{distance}{speed}=\frac{L-d}{c}+\frac{d}{v}$

Thanks Mr. rude man

 

[laurieke]

thank you so much for your help!