Solve

Guides

0

Question

Open in App

Solution

Verified by Toppr

As the cylinder is made of refractive index $$(-1)$$ and is placed in air of $$\mu =1$$, so when ray $$AB$$ is incident at B to cyliner, $${\theta}_{r}$$ will be negative ie., refractive ray will get reflected from normal. Similar thing happens at incidence point $$C$$ and angle of refraction and incident at $$B$$ and $$C$$ will be equal $$({\theta}_{r})$$ as $$OB=OC=R$$ and of refraction at $$C$$ is $${\theta}_{r}$$

$${\theta}_{1}=\left| { \theta }_{ i } \right| =\left| { \theta }_{ r } \right| =\left| { \theta }_{ r } \right| $$ as reflection takes place

The total deviation of out-coming ray from the incoming ray $$4{\theta}_{1}$$. Rays shall not reach the receiving plane if $$\cfrac { \pi }{ 2 } \le 4\theta \le \cfrac { 3\pi }{ 2 } $$ angles measured clockwise from the y axis

or $$\cfrac { \pi }{ 8 } \le \theta \le \cfrac { 3\pi }{ 8 } $$ (on dividing by 4 to all sides)

$$\sin { { \theta }_{ 1 } } =\cfrac { x }{ R } $$

$$\cfrac { \pi }{ 8 } \le \sin ^{ -1 }{ \cfrac { x }{ R } } \le \cfrac { 3\pi }{ 8 } \quad or\quad \cfrac { \pi }{ 8 } \le \cfrac { x }{ R } \le \cfrac { 3\pi }{ 8 } $$

Thus for light emitted from the source shall not reach the receiving plane if $$\cfrac { R\pi }{ 8 } \le x\le \cfrac { 3R\pi }{ 8 } $$

Was this answer helpful?

1