Question

The focal lengths of a thin lens for the red and violet colours are $f _{ R }$ and $f _{ v }$ respectively. The focal length $f$ of the lens and the dispersive power of its material for the mean colour are :

• A. $\frac{f_{R}+f_{v}}{2} ; \frac{f_{R}-f_{v}}{f}$
• B. $\sqrt{f_{R} f_{v}} ; \frac{f_{R}+f_{v}}{2}$
• C. $\sqrt{f_{R} f_{v}} ; \frac{f_{R}-f_{v}}{2}$
• D. $\frac{f_{R}+f_{v}}{2} ; \sqrt{f_{R} f_{v}}$

Answer 1

Answer: (C)

Focal length of the lens for mean colour (yellow)
$f =\frac{ f _{ R }+ f _{ v }}{2}$
Dispersive power of the material of lens,
$\omega=\frac{\mu_{ v }-\mu_{ R }}{\mu-1}$
Using lens formula,
$\frac{1}{ f }=(\mu-1)\left[\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right] \dots$(i)
$\left(\mu_{ v }-1\right)=\frac{1}{ f _{ v } \cdot x }$
where, $x=\frac{1}{R_{1}}-\frac{1}{R_{2}}$ and $\left(\mu_{R}-1\right)=\frac{1}{f_{R} \cdot x}$
Subtracting,
$\mu_{ v }-\mu_{ R }=\left(\frac{1}{ f _{ v }}-\frac{1}{ f _{ R }}\right) \frac{1}{ x }=\left(\frac{ f _{ R }- f _{ v }}{ f _{ R } f _{ v }}\right) \frac{1}{ x }$
$\mu_{ v }-\mu_{ R }=\frac{ f _{ R }- f _{ V }}{ f ^{2} \cdot x }$
where $f =\sqrt{ f _{ R } f _{ v }}$
From eq. (i), $\frac{1}{ f }=(\mu-1)\left[\frac{1}{ R _{1}}-\frac{1}{ R _{2}}\right]=(\mu-1) x$
Putting the value of $\frac{1}{f}$ in eqn. (ii), we get
$\mu_{ v }-\mu_{ R }=\frac{\left( f _{ R }- f _{ V }\right)(\mu-1) x }{ fx }$
or $\frac{\mu_{ v }-\mu_{ R }}{\mu-1}=\frac{ f _{ R }- f _{ v }}{ f }=\omega$