Physics - Geometrical Optics Question with Solution | TestHub

Let the refracting angle of the thin prism be $A$ and its refractive index be $\mu$.

Since the monochromatic ray is incident normally on the first face, the angle of incidence is $i_1=0^{\circ }$, and the angle of refraction is $r_1=0^{\circ }$.
The ray strikes the second face at an angle of incidence $r_2=A$.
For the ray passing through the prism, the angle of emergence $e$ is given by Snell's law:
$\sin e=\mu \sin A$
For a thin prism, angles are small, so $e\approx \mu A$.
The deviation $\delta$ of the emergent ray is:
$\delta =e-A=(\mu -1)A$
Given that $\delta =1.15^{\circ }$, we have:
$(\mu -1)A=1.15^{\circ }$ ... (1)

Now, consider the ray that is internally reflected from the second face.
The angle of incidence at the second face is $A$, so the angle of reflection is also $A$.
The reflected ray makes an angle of $2A$ with the normal to the first face (which is the direction of the incident ray).
This ray now strikes the first face from inside the prism at an angle of incidence $2A$.
Let the new angle of emergence be $e^{\prime }$. Using Snell's law at the first face:
$\sin e^{\prime }=\mu \sin (2A)$
For small angles, $e^{\prime }\approx \mu (2A)=2\mu A$.
This emergent ray makes an angle $e^{\prime }$ with the normal to the first face. Since the incident ray is along the normal, the angle between the emergent ray and the incident ray is $e^{\prime }$.
Given that this angle is $6.3^{\circ }$, we have:
$2\mu A=6.3^{\circ }$
$\mu A=3.15^{\circ }$ ... (2)

Substituting $\mu A=3.15^{\circ }$ into equation (1):
$\mu A-A=1.15^{\circ }$
$3.15^{\circ }-A=1.15^{\circ }$
$A=2.0^{\circ }$

Now, substituting the value of $A$ back into equation (2):
$\mu (2.0^{\circ })=3.15^{\circ }$
$\mu =\frac{3.15}{2.0}=1.575$

Answer: 1.575