An isosceles prism of angle 120^circ has a re | vedclass

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(D) The prism is isosceles with an apex angle of $120^\circ$. The base angles are $(180^\circ - 120^\circ)/2 = 30^\circ$.Let the incident rays be norm

Vedclass · Accepted Answer

Make an angle $2\{\sin^{-1}(0.72) - 30^\circ\}$ with each other

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(D) The prism is isosceles with an apex angle of $120^\circ$. The base angles are $(180^\circ - 120^\circ)/2 = 30^\circ$.Let the incident rays be normal to the first face. However,based on the geometry,the rays enter the prism and refract. Let the angle of incidence at the first surface be $i = 30^\circ$ relative to the normal.Applying Snell's law at the first surface: $1 \cdot \sin(30^\circ) = 1.44 \cdot \sin(r)$.$\sin(r) = \frac{0.5}{1.44} = \frac{50}{144} = \frac{25}{72} \approx 0.347$.Wait,looking at the provided solution geometry: The angle of incidence at the prism face is $30^\circ$ relative to the surface normal. Thus,$\sin(30^\circ) = 1.44 \sin(r) \Rightarrow \sin(r) = 0.5 / 1.44 = 0.347$. Actually,the provided solution uses $\sin(r) = 0.72$. This implies the angle of incidence was $i$ such that $\sin(i) = 1.44 	imes 0.72 = 1.036$,which is impossible. Re-evaluating the geometry: The angle of the prism is $120^\circ$. The rays are incident normally on the first face. The angle of refraction $r$ at the second face is $30^\circ$. Using Snell's law: $1.44 \sin(30^\circ) = 1 \sin(e) \Rightarrow \sin(e) = 1.44 	imes 0.5 = 0.72$.Thus,$e = \sin^{-1}(0.72)$. The deviation of each ray is $\delta = e - r = \sin^{-1}(0.72) - 30^\circ$.The total angle between the two emerging rays is $2\delta = 2\{\sin^{-1}(0.72) - 30^\circ\}$.

An isosceles prism of angle $120^\circ$ has a refractive index of $1.44$. Two parallel monochromatic rays enter the prism parallel to each other in air as shown. The rays emerging from the opposite faces

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