Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Ray Optics

Question:

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:

Options:

45cm, 12.5cm

90cm, 25cm

6cm, 12cm

25cm, 12.5cm

Correct Answer:

45cm, 12.5cm

Explanation:

For achromatism, $\frac{0.33}{f_1}+\frac{0.24}{f_2}=0$ or $33f_2 = −24f_1$

Again, for equivalent focal length F,

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$⇒\frac{1}{F}=\frac{1}{f_1}-\frac{33}{24f_1}=-\frac{3}{8f_1}$

or $\frac{1}{f_1}=-\frac{8}{150}$

and $\frac{1}{f_2}=+\frac{11}{150}$

Thus the lenses are concave (f1) and convex (f2). For the concave lens, from lens formula

$\frac{1}{f}=(μ-1)(\frac{1}{r_1}-\frac{1}{r_2})⇒-\frac{8}{150}=(μ-1)(\frac{1}{-r_1}-\frac{1}{+15})$

r1 = 45cm

Similarly for convex lens, $\frac{11}{150}=(1.5-1)(\frac{1}{15}-\frac{1}{-r_2})$

⇒ r2 = 12.5 cm