Chapter 43: Problem 15

Question: Prove that a ray incident oin the surface of a sheet of plate glass of thickness t emerges from the opposite face parallel to its incident direction but displaced sideways, as in Fig 30, Chap 43.

1.

Show that, for small angles of incidence $\theta$, this displacement is given by

$$x~=~t\theta {n-1\over n}$$

where n is the refractive index and $\theta$ is measured in radians.

2.

Calculate the displacement at a 10^o angle of incidence through a $1~{\rm cm}$ thich sheet of glass.

Solution: For small angles we have that $n_1\sin\theta_1~=~n_2\sin\theta_2$ becomes $n_1\theta_1~=~n_2\theta_2$. For small angles, the displacement of the ray in the absence of the glass is $X_1 = t\theta_1$, and the for the ray in the glass is $X_2=t\theta_2$, and therefore, we find the difference to be $x~=~X_1-X_2~=~t(\theta_1-\theta_2)~=~t\theta_1 ( 1 - {n_1\over n_2})$, which reproduces the required result when n₁ = 1 and n₂ = n. For part (b), one simply puts in the values of n = 1.55, $theta~=~0.17~{\rm rad}$, and $t~=~1~{\rm cm}$ to find $x~=~0.06~{\rm cm}$.