Prism And Dispersion

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A thin glass prism of refractive index 1.5 has a refracting angle of 6 degrees, and a physics student wants the angle of minimum deviation it produces.

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For a thin prism, NCERT shows that the angle of deviation is nearly independent of the incidence angle and is given by the simple relation $\delta=(n-1)A$, where $A$ is the small refracting angle and $n$ the refractive index. Substituting $n=1.5$ and $A=6^\circ$ gives $\delta=(1.5-1)	imes6^\circ=0.5	imes6^\circ=3^\circ$. This compact formula follows from the general prism deviation equation once the angles are small enough that sines can be replaced by the angles themselves in radians. The value $6^\circ$ is wrong because it drops the $(n-1)$ factor and equates deviation with the prism angle. The value $9^\circ$ is wrong because it uses $(n+1)$ instead of $(n-1)$, adding rather than subtracting one. The value $1.5^\circ$ mistakenly multiplies the refractive index by a quarter of the angle. As stated in NCERT Class 12, Chapter 9, thin prisms produce a deviation proportional to both the refracting angle and the excess of the index over unity. A magnitude check confirms the deviation is smaller than the prism angle for index 1.5, consistent with a mild bending of the ray.

Prism And Dispersion

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