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The objective of a telescope is an achromatic doublet of focal length 50cm. The refractive index of the glasses of the lenses used are 1.6 and 1.5 respectively for yellow light. The radius of curvature of the surfaces in contact is 15cm. If the dispersive powers of the glasses are 0.33 and 0.24, the radii of curvature of the other surfaces are:
Option: 1
$45 \mathrm{~cm}, 12.5 \mathrm{~cm}$

Option: 2
$90 \mathrm{~cm}, 25 \mathrm{~cm}$

Option: 3
$6 \mathrm{~cm}, 12 \mathrm{~cm}$

Option: 4
$25 \mathrm{~cm}, 12.5 \mathrm{~cm}$

Answers (1)

best_answer

For achromatism, $\mathrm{\frac{0.33}{f_{1}}+\frac{0.24}{f_{2}}=0 \,or\, \quad 33 f_{2}=-24 f_{1}}$

Again, for equivalent focal length $\mathrm{F}$ ,
$\mathrm{ \frac{1}{F} =\frac{1}{f_{1}}+\frac{1}{f_{2}}}$
$\mathrm{ \Rightarrow \frac{1}{F} =\frac{1}{f_{1}}-\frac{33}{24 f_{1}}=-\frac{3}{8 f_{1}}}$
$\mathrm{ \text { or } \frac{1}{f_{1}} =-\frac{8}{150}}$
$\mathrm{ \text { and } \frac{1}{f_{2}} =+\frac{11}{150}}$
Thus the lenses are concave $\mathrm{\left(f_{1}\right)}$ and convex $\mathrm{\left(f_{2}\right)}$ .For the concave lens, from lens formula

$\mathrm{\frac{1}{f}=(\mu-1)\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right) \Rightarrow-\frac{8}{150}=(\mu-1)\left(\frac{1}{-r_{1}}-\frac{1}{+15}\right)}$
$\mathrm{r_{1}=45 \mathrm{~cm}}$
Similarly for convex lens, $\mathrm{\frac{11}{150}=(1.5-1)\left(\frac{1}{15}-\frac{1}{-r_{2}}\right)}$

$\mathrm{\Rightarrow \quad r_{2}=12.5 \mathrm{~cm}}$

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