Question

Derivation of lateral displacement formula.

Answer 1

this is the derivation i got for uuu

Answer 2

Answer:

The lateral shift will be $\mathrm{t} \frac{\sin (t-R)}{\cos (R)}$

Explanation:

Here RT is lateral shift and RS  it is the emergent Ray

In triangle NQR,

Cos r is equal to $\frac{Q N}{Q R}$

$\Rightarrow Q R=\frac{Q N}{\cos r}$

Now we see in triangle QRT,

$\sin (\mathrm{i}-\mathrm{r})=\frac{R T}{QR}$

Applying value of QR from above equation

$\Rightarrow \sin (\mathrm{i}-\mathrm{r})=\frac{R T}{(Q N / \cos r)}$ on rearranging this equation we get

$\mathrm{RT}=\mathrm{QN} \frac{\sin (i-r)}{\cos r}$

Lateral shift will be $\mathrm{t} \frac{\sin (t-R)}{\cos (R)}$