A ray of light is incident at an angle of 9.3 | vedclass

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(C) For a small-angle prism,the relationship between the angle of incidence $i$,the angle of refraction $r_1$ at the first face,the angle of refractio

Vedclass · Accepted Answer

$1.55$

Vedclass · Answer

(C) For a small-angle prism,the relationship between the angle of incidence $i$,the angle of refraction $r_1$ at the first face,the angle of refraction $r_2$ at the second face,and the angle of emergence $e$ is given by:$A = r_1 + r_2$Given that the ray emerges normally from the second face,the angle of emergence $e = 0^{\circ}$,which implies that $r_2 = 0^{\circ}$.Since the refracting angle $A = 6^{\circ}$,we have $r_1 = A - r_2 = 6^{\circ} - 0^{\circ} = 6^{\circ}$.According to Snell's Law at the first face,$n_1 \sin(i) = n_2 \sin(r_1)$,where $n_1 = 1$ (air) and $n_2 = n$ (prism material).$1 \cdot \sin(9.3^{\circ}) = n \cdot \sin(6^{\circ})$.Using the small-angle approximation $\sin(	heta) \approx 	heta$ (in radians) or simply the ratio of angles for small values:$n = \frac{\sin(9.3^{\circ})}{\sin(6^{\circ})} \approx \frac{9.3}{6} = 1.55$.Thus,the refractive index of the material of the prism is $1.55$.

$A$ ray of light is incident at an angle of $9.3^{\circ}$ on one face of a small-angle prism with a refracting angle of $6^{\circ}$. If the ray of light emerges normally from the second face,the refractive index of the material of the prism is:

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