Question

Write a relation for the angle of deviation $\left(\delta \right)$ for a ray of light passing
through an equi-lateral prism in terms of the angle of incidence $\left(i\right)$, angle of emergence $\left(e\right)$ and angle of prism $\left(A\right)$.

• A. $\delta =(i-e)+A$
• B. $\delta =(i-e)-A$
• C. $\delta =(i+e)+A$
• D. $\delta =(i+e)-A$

Answer 1

The correct option is D $\delta =(i+e)-A$

The total angle of deviation is the sum of the angles it is bent by at the two surfaces i.e. ${\theta }_1=i-r_1$ and ${\theta }_2=e-r_2$ Thus,
$\delta =(i-r_1)+(e-r_2)=i+e-(r_1+r_2)$
The triangle made by top of the prism and the light ray must have a total internal angle of ${180}^o$. Thus,
$A+(90-r_1)+(90-r_2)=180$ i.e. $A=r_1+r_2$
Substituting in the equation for angle of deviation $\delta =i+e-A$
Therefore, $\delta +A=i+e$