Question 380: Ray Optics

Question

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index.

Accepted Answer

$$\mu $$ = $${{\sin \left( {{{{\delta _m} + A} \over 2}} ight)} \over {\sin \left( {{A \over 2}} ight)}}$$
= $${{\sin \left( {{{A + A} \over 2}} ight)} \over {\sin \left( {{A \over 2}} ight)}}$$
= $${{\sin A} \over {\sin \left( {{A \over 2}} ight)}}$$
= $${{2\sin \left( {{A \over 2}} ight)\cos \left( {{A \over 2}} ight)} \over {\sin \left( {{A \over 2}} ight)}}$$
= $$2\cos \left( {{A \over 2}} ight)$$
The angle of minimum deviation is given as
$$\delta $$
min
= i + e Ã¢ÂÂ A
for minimum deviation
$$\delta $$
min
= A and i = e then
2A = i + i
$$ \Rightarrow $$ i = A
i
min
= 0 = A
min
$$ \Rightarrow $$ $$\mu $$
min
= 2
i
max
= 90
0
= A
max
$$ \Rightarrow $$ $$\mu $$
max
= $$\sqrt 2 $$
$$ 	herefore $$ $$\mu $$ lies between 2 and $$\sqrt 2 $$.

Answer

lies between $$\sqrt 2 $$ and 1

Answer

lies between 2 and $$\sqrt 2 $$

Answer

is less than 1

Answer

is greater than 2

Ray Optics

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index.

Choose the correct answer: