Question

A ray of light is incident normally on the first refracting face of the 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 ${\sin }^{-1}\left(\frac{1}{2}\sec A/2\right)$
• B. Not emerge out of the prism
• C. Emerge at an angle of emergence ${\sin }^{-1}\left(\frac{1}{2}\sec A/4\right)$
• D. None of these

Answer 1

Answer: (A)

Emerge at an angle of emergence ${\sin }^{-1}\left(\frac{1}{2}\sec A/2\right)$

Given that 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

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

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