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Q. The refractive index of flint glass for red and blue colours are $1.664 \& 1.644 .$ What will be the dispersive power?

Ray Optics and Optical Instruments

Solution:

As we know
$\omega=\frac{\mu_{b}-\mu_{r}}{\mu-1}[\mu=$ mean refractive index $]$
$\mu=\frac{\mu_{ b }+\mu_{ r }}{2}$
$=\frac{1.664+1.644}{2}=1.654$
$\omega=\frac{\mu_{ b }-\mu_{ r }}{\mu-1}$
$\omega=\frac{1.664-1.644}{1.654-1}$
$=0.0305$