Question

For a prism, i $+e=A+\delta$ where
the symbols have their usual meanings.

• A. True
• B. False

Answer 1

The correct option is A True

From the figure, we may consider the triangle $\Delta MQR$. Here we get

$\measuredangle MQR+\measuredangle MRQ+\measuredangle QMR={180}^{\circ }$

$\Rightarrow \left({90}^{\circ }-r_1\right)+\left({90}^{\circ }-r_2\right)+\measuredangle A={180}^{\circ }$

$\Rightarrow \measuredangle A=r_1+r_2$ ----------$\left(1\right)$

Again from the figure, we can see that the incident ray after refraction is is bent by an angle say

${\theta }_1=i-r_1$

Similarly, the refracted ray is bent by an angle say

${\theta }_2=e-r_2$

Hence total deviation, $\delta ={\theta }_1+{\theta }_2$

$\Rightarrow \delta =i-r_1+e-r_2$

$\Rightarrow \delta =i+e-(r_1+r_2)$ ----------$\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we get,

$\delta =i+e-A$

$\Rightarrow i+e=A+\delta$

Hence the correct answer is option (A).