A prism of refractive index n_{1} and another | vedclass

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(B) For the light rays to pass through the interface $BC$ without bending (i.e., without refraction), the refractive indices of the two prisms must be

Vedclass · Accepted Answer

$600$

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(B) For the light rays to pass through the interface $BC$ without bending (i.e., without refraction), the refractive indices of the two prisms must be equal at that specific wavelength.Therefore, we set $n_{1} = n_{2}$:$1.2 + \frac{10.8 	imes 10^{-14}}{\lambda^{2}} = 1.45 + \frac{1.8 	imes 10^{-14}}{\lambda^{2}}$Rearranging the terms to solve for $\lambda$:$\frac{10.8 	imes 10^{-14}}{\lambda^{2}} - \frac{1.8 	imes 10^{-14}}{\lambda^{2}} = 1.45 - 1.2$$\frac{9 	imes 10^{-14}}{\lambda^{2}} = 0.25$$\lambda^{2} = \frac{9 	imes 10^{-14}}{0.25}$$\lambda^{2} = 36 	imes 10^{-14}$Taking the square root of both sides:$\lambda = 6 	imes 10^{-7} 	ext{ m}$Converting meters to nanometers $(1 	ext{ m} = 10^{9} 	ext{ nm})$:$\lambda = 6 	imes 10^{-7} 	imes 10^{9} 	ext{ nm} = 600 	ext{ nm}$

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