To determine the refractive index of a transparent plate of glass, a microscope is first focused on a tiny scratch in the upper surface, and the barrel position is recorded. Upon further lowering the microscope barrel by $\Delta d$, a focused image of the scratch is seen again. The plate thickness is $w$. What is the refractive index of the glass?

Clearly, this is due to the light from the scratch normal to the bottom surface that reflects and is transmitted through the upper surface and then travels to the microscope. This effective change in object distance is what counters the movement of the microscope to give a focused image again.

How though, do we calculate calculate the refractive index of the glass? My first thought was to equate the following quantities:

$$2w+(d-\Delta d)=d \\ 2w= \Delta d$$

Where $d$ was the original distance between the scratch and the microscope barrel. This is clearly wrong since the speed of light plays no part in it and therefore refractive index cannot be found from this equation.

My next thought was that perhaps the effective obeject distance is determined by the time light takes to travel. Using this logic, I came up with the following equation:

$$2\frac{w}{v}+\frac{d-\Delta d}{c}=\frac{\Delta d}{c}$$

Thereby equating the time light takes to travel from the scratch to the microscope before and after it is moved (NB $v$ is the speed of light in the glass). Strangely, this actually gave me the inverse of the correct answer.

The solution I have found here (see note below **) starts with the following equation:

$$\Delta d=\frac{2w}{n}$$

Where does this come from, and what physical reality does it describe? How does this relate to object distance? This gives the correct answer as it is stated in my textbook, so it is presumably correct.

** Note: The solution uses different symbols to the ones I have used. Take note especially of their different use of the letter $d$. Also, the question as originally formulated contains numerical values, which I have removed to make this question as general as possible.


I've realised that the solution arises from the following consideration of refraction:


In this case we know $v$ must be $-\Delta d$ because the scratch must appear to have moved below the surface of the glass by the same distance that the barrel of the microscope has moved.

Considering the reflection of the light at the bottom of the glass, the virtual image this reflection produces will be $w$ "beneath" the bottom surface, so therefore $u=2w$.

$$n_2 = 1, n_1=n \\ \Delta d = \frac{2w}{n}$$

The question can now be solved.


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