A parallel beam of white light falls on a con | vedclass

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(C) According to the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Since $\frac{1}{R_1} - \frac{1}{R_2}$

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$y > z > x$

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(C) According to the lens maker's formula,$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Since $\frac{1}{R_1} - \frac{1}{R_2}$ is constant for a given lens,we have $f \propto \frac{1}{\mu - 1}$.According to Cauchy's dispersion formula,the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ $(\mu \propto \frac{1}{\lambda})$.Therefore,light with a longer wavelength has a smaller refractive index,which results in a larger focal length $(f \propto \lambda)$.The wavelengths of the given colors are in the order $\lambda_R > \lambda_G > \lambda_B$.Consequently,the focal lengths follow the order $f_R > f_G > f_B$.Given that $x, y,$ and $z$ are the focal lengths for blue,red,and green light respectively,we have $x = f_B, y = f_R,$ and $z = f_G$.Since $f_R > f_G > f_B$,it follows that $y > z > x$.

$x$	$1$	$2$	$3$	$4$
$y$	$0.7111$	$0.7222$	$0.7333$	$0.7444$

$A$ parallel beam of white light falls on a convex lens. Images of blue,red,and green light are formed on the other side of the lens at distances $x, y,$ and $z$ respectively from the pole of the lens. Then:

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