Will the focal length of a lens for red light be more, same or less than that for blue light?
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Key concept: The refractive index depends on colour of light of wavelength of light.
Cauchy's equation: $\mu=\text{A}+\frac{\text{B}}{\lambda_2}+\frac{\text{C}}{\lambda^4}+\ ....$
As $\lambda_\text{red}>\lambda_\text{blue}\text{ hence }\mu_\text{red}<\mu_\text{blue}$
Hence parallel beams of light incidnt on a lens will be bent more towards the axis for blue light compared to red.
By lens maker's formula,
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
The refractive index for red light is less than that for blue light, $\mu_\text{red}<\mu_\text{blue}$
Hence $\frac{1}{\text{f}_\text{red}}<\frac{1}{\text{f}_\text{blue}}\Rightarrow\ \text{f}_\text{red}>\text{f}_\text{blue}$
Thus, the focal length for red light will ne greater than that for blue light.
Cauchy's equation: $\mu=\text{A}+\frac{\text{B}}{\lambda_2}+\frac{\text{C}}{\lambda^4}+\ ....$
As $\lambda_\text{red}>\lambda_\text{blue}\text{ hence }\mu_\text{red}<\mu_\text{blue}$
Hence parallel beams of light incidnt on a lens will be bent more towards the axis for blue light compared to red.
By lens maker's formula,
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{1}{\text{R}_1}-\frac{1}{\text{R}_2}\Big)$
The refractive index for red light is less than that for blue light, $\mu_\text{red}<\mu_\text{blue}$
Hence $\frac{1}{\text{f}_\text{red}}<\frac{1}{\text{f}_\text{blue}}\Rightarrow\ \text{f}_\text{red}>\text{f}_\text{blue}$
Thus, the focal length for red light will ne greater than that for blue light.
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