Question

show the deviation of less than 60°by a right angled prism.​

Answer 1

Explanation:

Total reflecting prisms are right angle isosceles prisms. They are based on the principle of total internal reflection. These prism are used to rotate light through 90 degrees or 180 degrees and are also used to obtain erect images of an inverted object.

When a ray of light incident normally on the surface of a right angled isosceles prism, it does not undergo refraction but strikes the surface AC at an angle 45 degrees, which is greater than the critical angle of glass, that is 42 degrees. The ray undergoes total internal reflection and strikes face BC normally and emerges out of it. The the ray is rotated through 90 degrees.

${\textstyle \implies \delta ={\frac {A}{2}}}{\textstyle \implies \delta ={\frac {A}{2}}}</p><p>$

Using the above formula,

{\textstyle {\frac {\sin \left({\frac {A+{\frac {A}{2}}}{2}}\right)}{\sin \left({\frac {A}{2}}\right)}}=1.4}{\textstyle {\frac {\sin \left({\frac {A+{\frac {A}{2}}}{2}}\right)}{\sin \left({\frac {A}{2}}\right)}}=1.4}

{\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\frac {1}{2}}{\frac {1}{\sqrt {2}}}}}{\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\frac {1}{2}}{\frac {1}{\sqrt {2}}}}}

{\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\sin 45^{\circ }}{\sin 30^{\circ }}}}{\textstyle \implies {\frac {\sin \left({\frac {3A}{4}}\right)}{\sin \left({\frac {A}{2}}\right)}}={\frac {\sin 45^{\circ }}{\sin 30^{\circ }}}}

{\textstyle \therefore A=60^{\circ }}{\textstyle \therefore A=60^{\circ }}