Question 253: Ray Optics

Question

A transparent solid cylindrical rod has a refractive index of $${2 \over {\sqrt 3 }}.$$ It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure.

The incident angle $$	heta $$ for which the light ray grazes along the wall of the rod is :

Accepted Answer

Applying Snell's law at $$Q$$
$$n = {{\sin {{90}^ \circ }} \over {\sin C}} = {1 \over {\sin C}}$$
$$	herefore$$ $$\sin C = {1 \over n} = {{\sqrt 3 } \over 2}$$
$$	herefore$$ $$C = {60^ \circ }$$
Applying Snell's Law at $$P$$
$$n = {{\sin 	heta } \over {\sin \left( {90 - C} ight)}}$$
$$ \Rightarrow \sin 	heta  = n 	imes \sin \left( {90 - C} ight);$$ from $$(1)$$
$$ \Rightarrow \sin 	heta  = n\,\cos C$$
$$	herefore$$ $$	heta  = {\sin ^{ - 1}}\left[ {{2 \over {\sqrt 3 }} 	imes \cos {{60}^0}} ight]$$
or, $$	heta  = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} ight)$$

Answer

$${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{$\scriptstyle {\sqrt 3 }$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{$\scriptstyle 2$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle {\sqrt 3 }$}}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$

Answer

$${\sin ^{ - 1}}\left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \right)$$

Ray Optics

A transparent solid cylindrical rod has a refractive index of ${2 \over {\sqrt 3 }}. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle \theta $ for which the light ray grazes along the wall of the rod is :

Choose the correct answer: