Question

A ray of light is incident normally on the first refracting face of the isosceles prism of refracting angle A. The ray of light comes out at grazing emergence. If one half of the prism (shaded position) is knocked off, the same ray will :

• A. emerge at an angle of emergence $si{n}^{-1}\left(\frac{1}{2}secA/2\right)$
• B. not emerge out of the prism
• C. emerge at an angle of emergence $si{n}^{-1}\left(\frac{1}{2}secA/4\right)$
• D. None of these

Answer 1

The correct option is A emerge at an angle of emergence $si{n}^{-1}\left(\frac{1}{2}secA/2\right)$

Given: A ray of light is incident normally on the first refracting face of the isosceles prism of refracting angle A. The ray of light comes out at grazing emergence.

To find the angle emergence angle of the same if one half of the prism is knocked off

Solution:

Case (i): when ray of light is incident normally on first refracting face of the isosceles prism. (refer fig(i))

Let the ray X is incident normally on first refracting face, AB of the isosceles triangle ABC.

As it it given refracting angle, $\angle XYZ=A$

Hence $\angle XYA=(90-A)$

In right angled traingle $AOY$

${180}^{\circ }=\angle AOY+\angle OYA+\angle YAO⟹\angle YAO=180-90-(90-A)=A.........\left(i\right)$

And as the comes out at grazing emergence hence angle of emergence in this case is ${90}^{\circ }$

So we know,

${\mu }_1\times \sin i={\mu }_2\times \sin e........\left(ii\right)$.

Here ${\mu }_1={\mu }_p$ refractive index of prism,

${\mu }_2={\mu }_a=1$ as it is refractive index of air,

angle of incidence $i=A$ (given) and

angle of emergence $e={90}^{\circ }$ (given).

Therefore, by substituting these values in equation (ii), we get

${\mu }_p\times \sin A=1\times \sin 90⟹{\mu }_p=\frac{1}{\sin A}.........\left(iii\right)$

Case (ii): when half of the prism is knocked off (refer fig(ii))

Then $\angle YAO=\frac{A}{2}$

In right-angled triangle,

${180}^{\circ }=\angle AOY+\angle OYA+\angle YAO⟹\angle OYA=180-90-\left(\frac{A}{2}\right)⟹\angle OYA=90-\frac{A}{2}$

And angle of incidence, $\angle XYZ=i=90-(90-\frac{A}{2})=\frac{A}{2}$

So the equation (ii), in this case will become

${\mu }_p\times \sin \left(\frac{A}{2}\right)={\mu }_a\times \sin e⟹\frac{1}{\sin A}\times \sin \left(\frac{A}{2}\right)=\sin e$

$\sin e=\frac{\sin \left(\frac{A}{2}\right)}{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}$ [$\because \sin A=2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)$]

Therefore,

$\sin e=\frac{\sec \left(\frac{A}{2}\right)}{2}$ [$\because \sec A=\frac{1}{\cos A}$]

Hence incident ray will emerge at angle of emergence

$e={\sin }^{-1}\left\{\frac{1}{2}\sec \left(\frac{A}{2}\right)\right\}$