Question

A telephoto combination consists of convex lens of focal length 30 cm and a concave lens of focal length 15 cm, the separation between two lenses is 28.4 cm. Determine where should be the photographic plate placed in order to photograph an object 10 m in front of the first lens?

• A. 3 cm
• B. 15 cm
• C. 6 cm
• D. 9 cm

Answer 1

The correct option is A 3 cm

u = -1000 cm

$f_1$ = 30 cm

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}+\frac{1}{1000}=\frac{1}{30}$

$\frac{1}{v}=\frac{1}{30}-\frac{1}{1000}$

$v$ = 30.927 cm

$u_2=v_1-28.4$

$u_2=+2.527$ cm

So, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{-15}+\frac{1}{2.527}$

$v=3.0391cm$

hence the screen should be placed at 0.0391 cm behind concave lens.

#

chromatic aberration = Dispersing Power $\times$ f

Dispersive Power = $\frac{{\mu }_v-{\mu }_r}{{\mu }_y-1}$

= $\frac{1.66-1.62}{1.64-1}$

D = 0.0625 cm

chromatic abberation = 0.625 cm